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English Pages [290] Year 2010
ELEMENTS OF POINTSET TOPOLOGY
Carlos Prieto∗
∗
Instituto de Matem´aticas, UNAM
2010
Date of version: October 28, 2009
Preface
Nowadays one speaks more and more about the specialization in modern science. Even though this statement is valid up to a certain point, one might say that a characteristic of science today is the every time greater interaction among the various disciplines that conform it. Similarly to to what happens in science in general, in each discipline one pursues a broader relationship among the different fields that conform it. In mathematics, for instance, one expects from a differential geometer or from a function theorist a much wider common knowledge than the one required one half century ago. This happens because of the ubiquity that some mathematical concepts show more and more. One of these mathematical concepts is that of a topological space, that includes everything related with “nearness”, “continuity”, “neighborhood”, “deformation”, et cetera. Topology has been for many years one of the most important and influential fields in modern mathematics. Its origins date back over some centuries, but it was doubtless Poincar´e who imprinted the great impetus that topology gained throughout the twentieth century. There are other great names among those who created pointset topology, whose existence has been justified by the great progress of algebraic topology. On the one hand, the effectiveness of pointset topology, more than due to deep theorems, it rests in the first place on its conceptual simplicity and on its convenient terminology, because in a sense it establishes a link between abstract, not very intuitive problems, and our ability to visualize geometric phenomena in space. This intellectual ability to grasp what is going on in 3dimensional space, that through topology allows us to delve into mathematical thinking and into the world of abstract objects, is very independent of abstraction and logical thinking. This reinforcement of our mathematical talent is probably the deepest cause of the effectiveness and the simplicity of the topological methods. As many of the basic mathematical branches, topology has an intricate history. If we mark the start of topology at the point when the conceptual system of pointset topology was established, then we have to refer to Felix Hausdorff’s book Grundz¨ uge der Mengenlehre (Foundations of Set Theory), Leipzig, 1914, in whose Chapter 7 “Point Sets in General Spaces”, he establishes the most important and basic concepts in pointset topology. Already in 1906, in his paper Sur
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quelques points du calcul fonctionnel (On some Topics of Functional Calculus), Maurice Fr´echet introduced the concept of metric space and he tried to establish the concept of topological space, by giving an axiomatic approach to the concept of convergence. What Fr´echet really created were the topological foundations of functional analysis. But, of course, the history dates further back to the times when the effervescence of geometry was taking place during the nineteenth century. At the beginning of that century there was the classical idea that geometry was the mathematical ambit, where the concepts of the physical space developed. Towards the end of that century, as it was shown by Felix Klein in his Erlanger Programm (Erlangen Program. Comparative Considerations about New Geometric Investigations), the projection went much farther than the physical space and it even started to consider such abstract spaces as the nmanifolds, the projective spaces, the Riemann surfaces, or even the function spaces. Among the decisive works for the emergence of topology, one finds the monumental work of Georg Cantor. He established the bases on which the abstract concept of a topological space is formulated as “a set furnished with a collection of subsets such that...” Indeed, already in 1870, Cantor had shown that if two Fourier series converge pointwise and have the same limit, then they must have the same coefficients. Cantor himself improved this result in 1871 by showing that the coincidence of the coefficients can equally be achieved by requiring pointwise convergence or equality of the limits, up to a finite set in the interval [0, 2π]. In 1872 he analyzed certain infinite subsets, up to which his statement remains valid. It was then, when he introduced his famous Cantor set, that being “only” a subset of an interval, it is topologically not only a very interesting object, but of great importance in several branches of mathematics. The problem of deciding if two spaces are homeomorphic or not is no doubt the central problem in topology. It was not until the creation of algebraic topology that it was possible to give a reasonable answer to such a problem. Now it is not only because of its conceptual simplicity and its adequate symbology, but thanks to the powerful tool provided by algebra and its most convenient functorial relationship to topology that this effectiveness is achieved. The analytic description of dynamical systems in classical mechanics represented the first step towards the necessity to create a geometrical language in dimensions higher than the usual ones. Already in the eighteenth century, Lagrange, in his M´ecanique analytique (Analytical Mechanics) Paris, 1788, had considered the possibility of grasping a fourth dimension. It was Riemann, in his famous Ha¨ bilitationsvortrag: Uber die Hypothesen, welche der Geometrie zu Grunde liegen (On the Hypotheses underlying Geometry), G¨ottingen, 1854, who presented the
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first ideas on the geometry of manifolds. Lagrange himself, in his Le¸cons sur le Calcul des Fonctions (Lessons on Functional Calculus), Paris, 1806, introduced the concept of perturbation, or homotopy, of curves in variational calculus problems to detect certain minimal curves. What we now know as algebraic topology was probably started with Henri Poincar´e’s Analysis Situs, Paris, 1895, and its five Compl´ements (Complements), Palermo 1899, London 1900, Paris 1902, Paris 1902, and Palermo 1904. In the first, he notices that “geometry is the art of reasoning well with badly made figures.” And further says: “Yes, without doubt, but under one condition. The proportions of the figures might be grossly altered, but their elements must not be interchanged and must preserve their relative situation. In other terms, one does not worry about quantitative properties, but one must respect the qualitative properties. That is to say precisely those, which are the concern of Analysis Situs.” Indeed, other works of Poincar´e contain as much interesting topology as the ones just referred to. This is the case for his memoir on the qualitative theory of differential equations, that includes the famous formula of the Poincar´e index. This formula describes in topological terms the famous Euler formula, and constitutes one of the first steps in algebraic topology. In the works mentioned, Poincar´e considers already maps on manifolds such as, for instance, vector fields, whose indexes determine the Euler characteristic in his index formula. It is Poincar´e too who generalizes the question on the classification of manifolds having in mind the classification of the orientable surfaces considered by Moebius in his Theorie der elementaren Verwandtschaft (Theory of elementary relationship), Leipzig, 1863. This classification problem was also solved by Jordan in Sur la d´eformation des Surfaces (On the deformation of surfaces), Paris, 1866, who, by classifying surfaces solved an important homeomorphism problem. Jordan also studied the homotopy classes of closed paths, that is, the first notions of the fundamental group, inspired by Riemann, who already had analyzed the behavior of integrals of holomorphic differential forms and therewith the concept of homological equivalence between closed paths. Of course, Cantor, Fr´echet, Klein, Hausdorff, Riemann, Jordan, Moebius, and Poincar´e are not the only architects of the basic concepts of topology. All this history is in itself the object of another text. Doubtless, the text [10] edited by I.M. James is an excellent reference in that direction. This book has the purpose of presenting the topics of pointset topology, which from my own point of view, are basic for an undergraduate student, who is interested in this area or affine areas in mathematics. The design of the text is as follows. We start with a small rather motivating Chapter 1, followed by six substantial chapters, each of which is divided into
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several sections that are distinguished by their double numbering (1.1, 1.2, 2.1, . . . ). Definitions, propositions, theorems, remarks, formulas, exercises, etc., are designated with triple numbering (1.1.1, 1.1.2, . . . ). Exercises are an important part of the text, since many of them are intended to carry the reader further along the lines already developed, in order to prove results that are either important by themselves or relevant for future topics. Most of these are numbered, but occasionally they are identified inside the text by the use of italics (exercise). The book starts considering metric spaces to arrive to the abstract properties of their open sets. They lead us to the abstract concept of a topological space. Then it studies continuity as the fundamental property expected from any function between topological spaces. Further, one studies several conditions which are additional to the very axioms, that guarantee useful and convenient properties of the spaces. Special emphasis is put on compact spaces, because this concept has a special importance in several applications of topology. We finish with the ˇ metrizability theorems and the Stone–Cech compactification. Along the book, we remark the universal properties that the many of the given constructions have. Particularly the universal properties that characterize the topological sum and the topological product. We also give the universal properties of the identifications, ˇ the Stone–Cech compactification, et cetera. A section is devoted to the important topic on limits and colimits of topological spaces, and one more to the compactly generated spaces, since they play an important role in algebraic topology. The book is intended to be used in a onesemester course in the higher undergraduate level. To do this, one may read Chapters 1–3. In Chapter 4, one can leave out Section 4.6, which refers to limits and colimits. In Chapter 5, one can omit Section 5.6, which is devoted to nets. In Chapter 6, Section 6.7, which handles compactly generated spaces, can be left out. Finally in Chapter 7, Sections 7.5 and 7.6, which deal with paracompact spaces and with the interrelations among several properties, can be omitted. This shortcut fulfills the purpose of the book of starting with metric spaces and, after adding adequate conditions to general topological spaces, coming back to metrizable spaces. On the other hand, the omitted sections can be left to the students to develop different projects. In particular, Section 6.7, besides the compactly generated spaces, describes briefly the class of kspaces, which represents a very interesting project for good students. At this point I want recognize the big impact in this book of all experts that directly or indirectly have influenced my education as a mathematician and as a topologist. At the UNAM, Guillermo Torres and Roberto V´azquez were decisive. Later on, in my doctoral studies in Heidelberg, Germany, I had the privilege of receiving directly the teaching of Albrecht Dold and Dieter Puppe. Indirectly, I was influenced by some German topology books, among which I owe a mention to that of Klaus J¨anich [11] –whose effects are clearly reflected, mainly in this
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preface–. Horst Schubert’s influence [16] is clear all through the book.
Mexico City, Mexico Summer 2009
Carlos Prieto
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Contents
Preface
iii
Introduction
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1 Metric spaces
1
1.1
Euclidean spaces . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.3
Neighborhoods and open sets . . . . . . . . . . . . . . . . . . . .
6
1.4
Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
1.5
Pseudometric spaces . . . . . . . . . . . . . . . . . . . . . . . . .
12
2 Topological spaces
17
2.1
Basic definitions: open sets and neighborhoods . . . . . . . . . .
17
2.2
Closed sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
2.3
Other basic concepts . . . . . . . . . . . . . . . . . . . . . . . . .
28
2.4
Neighborhood bases . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.5
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
2.6
Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
3 Comparison of topologies
39
3.1
Comparison of topologies . . . . . . . . . . . . . . . . . . . . . .
39
3.2
Intersection of topologies . . . . . . . . . . . . . . . . . . . . . . .
41
3.3
Supremum of a family of topologies . . . . . . . . . . . . . . . . .
42
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3.4
Basis of a topology . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Generating topological spaces
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53
4.1
Induced topology . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.2
Identification topology . . . . . . . . . . . . . . . . . . . . . . . .
60
4.3
Topological product . . . . . . . . . . . . . . . . . . . . . . . . .
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4.4
Topological sum
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. . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Limits and colimits
91
5.1
Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.2
Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.3
Colimits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5.4
Special constructions . . . . . . . . . . . . . . . . . . . . . . . . .
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5.5
Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6 Connectedness
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6.1
Connected spaces . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6.2
Locally connected spaces . . . . . . . . . . . . . . . . . . . . . . .
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6.3
Pathconnected spaces . . . . . . . . . . . . . . . . . . . . . . . .
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6.4
Locally pathconnected spaces . . . . . . . . . . . . . . . . . . . .
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7 Filters
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7.1
Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.2
Cluster points . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.3
Ultrafilters
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7.4
Filters and functions . . . . . . . . . . . . . . . . . . . . . . . . .
149
7.5
Filters and products . . . . . . . . . . . . . . . . . . . . . . . . .
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7.6
Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8 Compactness 8.1
Compact sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
163 163
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8.2
Compactness and countability . . . . . . . . . . . . . . . . . . . .
175
8.3
The Alexandroff compactification . . . . . . . . . . . . . . . . . .
183
8.4
Proper maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
190
8.5
Compactopen topology . . . . . . . . . . . . . . . . . . . . . . .
195
8.6
The exponential law . . . . . . . . . . . . . . . . . . . . . . . . .
199
8.7
Compactly generated spaces . . . . . . . . . . . . . . . . . . . . .
203
8.8
kSpaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
213
9 Other separability axioms
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9.1
Normal spaces
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
217
9.2
Completely regular spaces . . . . . . . . . . . . . . . . . . . . . .
227
9.3
ˇ The Stone–Cech compactification . . . . . . . . . . . . . . . . . .
231
9.4
Metrizable spaces . . . . . . . . . . . . . . . . . . . . . . . . . . .
235
9.5
Paracompact spaces . . . . . . . . . . . . . . . . . . . . . . . . .
238
9.6
Interrelations among topological properties . . . . . . . . . . . .
249
References
253
Symbols
271
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Introduction
The object of this book is to study topology. But, what is topology? This is not an easy question to answer. Trying to define a branch of mathematics in a concise sentence is complicated. However, as an approximation, we can say that topology is the branch of mathematics that studies continuous deformations of geometric objects. One of the purposes of topology is to classify objects, or at least, to give methods in order to distinguish between objects that are not homeomorphic. Namely to decide between objects that cannot be obtained from each other through a continuous deformation. Topology also provides techniques to study topological structures in objects that arise in very different fields of mathematics. Those concepts such as “deformation, “continuity” and “homeomorphism” are fundamental and will be precisely defined throughout the text. However, although we have not defined these concepts yet, we introduce in what follows several examples, that intuitively illustrate those concepts. Figure 0.1 shows three topological spaces, namely a spherical surface with its poles deleted, another sphere with its polar caps removed (including the polar circles), and a cylinder with its top and bottom removed (including the edges). These three objects can clearly be deformed one to the other. Topology would not distinguish them.
Figure 0.1 A sphere with no poles, a sphere with no polar caps and circles, and a cylinder with no edges Figure 0.2 shows the surface of a torus (a “doughnut”) and a spherical surface xiii
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with a handle attached. Each one of these objects is clearly a deformation of the other. However, it is intuitively clear that the topological object shown in three forms in Figure 0.1, i.e. the 2sphere with no poles and the other two, cannot be deformed to the topological object shown in two forms in Figure 0.2.
Figure 0.2 A torus and a sphere with one handle Being a little more precise, we shall say that two objects (topological spaces) will be homeomorphic when there exists a onetoone correspondence, which maps points that are close in one of the objects to points that are close in the other. We can add to the list of those spaces, mentioned above, which are not homeomorphic to each other, the following examples: (a) Take N = {0, 1, . . . , n − 1}, M = {0, 1, . . . , m − 1}, n < m. Considered as topological spaces in any possible way, they will never be homeomorphic, since a necessary condition in order for two spaces to be homeomorphic, is that they have, as sets, the same cardinality, i.e the same number of elements. (b) The same argument of (a) shows that a onepoint set cannot be homeomorphic to an interval. (c) More sophisticated arguments are required to show that a closed interval is not homeomorphic to a cross; that is, the topological spaces depicted in the upper part of Figure 0.3 are not homeomorphic to each other. A way of deciding this might be the following: Whatever point we delete from the interval decomposes it in at most two connected portions. However, there is one point in the cross that when deleted, it decomposes the cross into four pieces (components). For that reason, no point in the first space can exist that would correspond to this special point in the second space under a homeomorphism. Therefore, they cannot be homeomorphic. (d) The surface of a torus is not homeomorphic to that of a sphere. This might be shown if we observe that a circle (a simple closed curve) can be drawn on the surface of the torus that cannot be continuously contracted to a point.
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Figure 0.3 An interval and a cross are not homeomorphic However, it is very clear that any circle drawn in the surface of a sphere can be deformed to a point, as can be appreciated in Figure 0.4. Another way of seeing this might be observing that any circle on the sphere is such that when deleted, the sphere is decomposed in two regions, while the circle on the torus does not have this property. (In other words, the famous Jordan curve theorem holds on the sphere, while it does not hold on the torus.)
Figure 0.4 Any loop can be contracted on the sphere and a loop that does not contract on the torus (e) The Moebius band that can be obtained from a strip of paper twisting it one half turn and then gluing it along its ends, is not homeomorphic to the trivial band obtained from a similar strip by gluing its ends without twisting it (see Figure 0.5). The argument for showing this can be similar to that used in Example (d), namely there is a circle on the Moebius band, that if removed from the band, does not disconnect it (it can be cut with scissors along the equator
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Figure 0.5 The Moebius band and the trivial band and would not fall apart in two pieces). However, in the trivial band any circle parallel to and different from the edges (or any other circle different from the edges), when removed, decomposes the band in two components (see Figure 0.6). The first exercises for the reader are the following: (f) Take the Moebius band and cut it along the equator. What space do you obtain? Will it be a Moebius band again? Or maybe will it be the trivial band?
Figure 0.6 The Moebius band is not homeomorphic to the trivial band (g) Similarly to the given construction of the Moebius band we may take a paper strip and glue its ends, but this time after twisting a full turn. Will this space be homeomorphic to the Moebius band? Or will this space be homeomorphic to the trivial band? What is the relationship between this space and that of (f)? (see Figure 0.6.) One of the central problems in topology consists, precisely, in studying topological spaces in order to be able to distinguish between them. In all the previous examples every time we have decided that two spaces are not homeomorphic, it has been on the base of certain invariants that can be assigned to them. For
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instance, in (a) this invariant is the cardinality, in (c) it is the number of components obtained after deleting a point, and in (d) and (e) it is the number of components obtained after removing a circle. One of the goals pursued by topology is to assign to each space invariants that are relatively easy to compute and allow to distinguish among them. When we mentioned the intuitive concept of homeomorphism, we used the intuitive concept of nearness of two points, that is, we talked about the possibility of deciding if a point is close to a given point, or equivalently if it is in a neighborhood of the given point. In the first chapters of this book we shall make this concept precise. A knot K is is a simple closed curve in the 3space, i.e. it is the image k(S1 ) under a “decent” inclusion k : S1 ,→ R3 of the unit circle in the plane into the 3dimensional Euclidean space as intuitively shown in Figure 0.7.
Figure 0.7 A knot in 3space Knot theory is an important field of mathematics with striking applications in several branches of science. The central problem of the theory consists in determining when two given knots are equivalent; that is, when is it possible to deform inside the space one knot to the other without tearing it apart. A result in the theory proved by Gordon and Luecke [8] states that a knot K is determined by its complement, that is, that two knots K and K 0 are equivalent if and only if their complements R3 − K and R3 − K 0 are homeomorphic. In other words, they transformed the problem of classifying knots into a homeomorphism problem of certain open sets in R3 .
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Chapter 1
Metric spaces
A very rich source of topological examples are the metric spaces, since, in a very natural way, they have the fundamental topological properties. In this chapter we study briefly the basic concepts of the theory of metric spaces. We start by discussing Euclidean spaces and their subspaces.
1.1
Euclidean spaces
In order to become familiar with the general notation of this book, in this section we present a series of examples of “canonical” topological spaces that play a very important role in topology, as well as in other branches of mathematics. The symbols R and C will denote, as usual, the spaces of the real and the complex numbers. If n ≥ 1, then Rn will be the Euclidean space of dimension n, that is, Rn = {x = (x1 , . . . , xn )  xi ∈ R, i = 1, . . . , n} with its usual operations as a vector space: if x, y ∈ Rn y r ∈ R, then x + y = (x1 + y1 , . . . , xn + yn ), x − y = (x1 − y1 , . . . , xn − yn ) and rx = (rx1 , . . . , rxn ), and the norm is given p by x = x21 + · · · + x2n . One defines the distance between two points simply by y − x. R0 represents the Euclidean space consisting of only one point, the 0, that can be seen as a subspace of the space Rn . There is a canonical identification Rn × Rm = Rn+m given by ((x1 , . . . , xn ), (y1 , . . . , ym )) = (x1 , . . . , xn , y1 , . . . , ym ). Similarly, Rn can be canonically seen as a subspace of Rn+1 identifying it with the subspace Rn × 0. There is also a canonical identification of R2 with the complex √ plane C by setting (x, y) = x + iy, where i = −1. 1.1.1 Definition. Let n ≥ 0. Consider the following spaces:
R+ = {x ∈ R  0 ≤ x}, the nonnegative halfline. Bn = {x ∈ Rn  x ≤ 1}, the unit ball of dimension n, or unit nball. Sn−1 = {x ∈ Rn  x = 1}, the unit sphere of dimension n − 1, or unit (n − 1)sphere. ◦
n
B = {x ∈ Rn  x < 1}, the unit cell or open unit ball of dimension n or unit ncell. 1
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1 Metric spaces
I n = {x ∈ Rn  0 ≤ xi ≤ 1, 1 ≤ i ≤ n}, the unit cube of dimension n. ∂I n = {x ∈ I n  xi = 0 or 1 for some i}, the boundary of I n in Rn . I = I 1 = [0, 1] ⊂ R, the unit interval. 1.1.2 Exercise. Prove the equality
S1 = {z ∈ C  z = 1} , that is, the equality
S1 = {e2πit ∈ C  t ∈ I} .
1.2
Metric spaces
In the usual metric on Rn , that we defined above, one considers the concept of open set. The concept of a metric and its associated concept of open set can be generalized to any set furnished with a function that behaves analogously to the distance function of Rn . This allows to study a series of properties of those sets with a metric, that are common to those properties of the Euclidean spaces and any of their subspaces. For instance, the concept of a metric is closely related to the concept of convergence of sequences. Frequently there are more general concepts, as that of convergence of a sequence of functions, which require a more general set up than the elementary concept of a Euclidean space. The purpose of this section is to establish an axiomatic system that includes the concepts of convergence, open sets, continuity, etc., that are familiar to us from elementary analysis. 1.2.1 Definition. A metric space consists of set X and a function d : X × X −→ R+ , called metric, or distance that satisfies the following axioms: (M1) d(x, y) = 0 ⇔ x = y. (M2) d(x, y) = d(y, x)∀x, y ∈ X. (M3) d(x, y) ≤ d(x, z) + d(y, z), ∀x, y, z ∈ X. This last inequality is known as the triangle inequality (see figure 1.1).
3
1.2 Metric spaces
x d(x, y)
y
d(x, z) d(y, z) z
Figure 1.1 The triangle inequality 1.2.2 Examples. The following are metric spaces: (a) X = R, d(x, y) = y − x. (b) X = Rn , d(x, y) = y − x. (c) X ⊆ Rn , d(x, y) = y − x. X is called metric subspace of Rn . Examples of ◦
n
metric subspaces of Rn are the following: Bn , B , Sn−1 , I n , ∂I n . From here on, we shall consider these subspaces as metric spaces with this concept. (d) Let X be any nonempty set and let d : X × X → R+ be given by ½ 0, si x = y; d(x, y) = 1, si x 6= y . This is clearly a metric which will be called the discrete metric on X. (e) X = Rn , d(x, y) = max{xi − yi   i = 1, . . . , n}. (f) X = {x = (xi )  xi ∈ R, i ∈ N,
P∞
2 i=1 xi
< ∞}, v u∞ uX d(x, y) = t (yi − xi )2 i=1
This space of sequences of real numbers is usually denoted in functional analysis by `2 and is called the (real) Hilbert space. (g) X = {x : I −→ R  x is a continuous function}, s Z 1 d(x, y) = (x(t) − y(t))2 dt . 0
(h) X = {x : I −→ R  x is a continuous function}, d(x, y) = m´ax{x(t) − y(t)  t ∈ I}
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1 Metric spaces
(i) X = {x : I −→ R2  x is a continuous function, x(0) = x(1)}, d(x, y) = m´ax{x(t) − y(t)  t ∈ I}. Observe that in the last example two different functions may have the same image, but however have a nonzero distance between them. For instance, the functions x, y : I −→ R2 given by x(t) = (cos 2πt, sin 2πt) , y(t) = (cos 4πt, sin 4πt) satisfy d(x, y) = 2; however, the image of each of them is S1 . It is not elementary to prove that the function d in each of the previous examples is indeed a metric. In some cases, certain important (and famous) inequalities are needed. The reader can see [14] or [2], for instance. Notice that example (d) shows that any set can be given a metric. On the other hand, examples (b) and (e) or (g) and (h) show that the same set may have different nondiscrete metrics. In what follows, Rn will always denote the metric space of example (b). 1.2.3 Remark. In the definition of a metric, it is enough to ask that d is a realvalued function and that axioms (M1) and (M3) hold. Axiom (M2) is a consequence of the other two as follows. If we take z = x, then we have by (M3) that d(x, y) ≤ d(x, x) + d(y, x) = d(y, x), where the last equality holds by (M1). Analogously, d(y, x) ≤ d(x, y). Hence d(x, y) = d(y, x) and so (M2) holds. Moreover, if we take y = x, then by (M1) and (M3) we have 0 = d(x, x) ≤ d(x, z) + d(x, z) = 2d(x, z). Thus d(x, z) ≥ 0 for any x, z and the function d is nonnegative. 1.2.4 Proposition. Let (X, d) be a metric space and consider any subset Y ⊆ X. Then the restriction d0 = dY ×Y : Y × Y −→ R+ is a metric on Y . t u The metric space Y with the restricted metric will be called a metric subspace of X. 1.2.5 Exercise. In each of the following cases, prove that the given function d : X × X ∈ R+ is a metric. (a) In X = Rn take d(x, y) =
n X i=1
xi − yi .
5
1.2 Metric spaces
(b) In X = {x : I −→ Rn  x is a continuous function} take Z 1 d(x, y) = x(t) − y(t)dt. 0
(c) In X = {x = (xi )  xi ∈ R, i ∈ N, xi → 0} take d(x, y) = sup{xi − yi   i ∈ N}. (d) In X = {x = (xi )  xi ∈ R, i ∈ N, d(x, y) =
P∞
p i=1 xi 
Ã∞ X
< ∞}, for some p ≥ 1 take !1
p
xi − yi p
.
i=1
(Hint: Use Minkowski’s inequality [2].) (e) Take X as in (d) and assume 0 < p < 1. Let d be given by the same formula of (d). (f) If di : Xi × Xi −→ R is a metric on a set Xi , i = 1, . . . , n, take X = and take n X d(x, y) = di (xi , yi ) .
Qn
i=1 Xi
i=1
(g) If di : Xi × Xi −→ R is a metric on a set Xi , i = 1, . . . , n, take X = and for some p ≥ 1 take d(x, y) =
Ã n X
Qn
i=1 Xi
!1
p
di (xi , yi )p
.
i=1
The metric spaces of (d) and (e) are usually denoted by `p . Th metric of (f) is called product metric. 1.2.6 Definition. Let X be a real vector space. A norm on X is a function that maps each x ∈ X to a real number kxk ∈ R+ , called the norm of x, such that the following conditions hold: (No1) kxk = 0 ⇔ x = 0, (No2) krxk = r · kxk, r ∈ R, (No3) kx + yk ≤ kxk + kyk. A vector space X furnished with a norm is called a normed vector space.
6
1 Metric spaces
1.2.7 Exercise. Prove the following statements. (a) In a normed vector space X the function d(x, y) = kx − yk defines a metric. (b) In Rn each of the following functions defines a norm: (i) kxk = m´ax{xi   i = 1, . . . , n} P (ii) kxk = ( ni=1 xi p )1/p , p ≥ 1. The metric associated to (i) is that of 1.2.2(e) and the metric associated to Q (ii) is that of 1.2.5(g) if Xi = R and we take Rn = ni=1 R.
1.3
Neighborhoods and open sets
As we have already said, in a metric space it is possible to talk about neighborhoods of a point. 1.3.1 Definition. Take x ∈ X, ε ∈ R, ε > 0. We define the open ball with center x and radius ε by Bε (x) = {y ∈ X  d(x, y) < ε} . Furthermore we say that U ⊂ X is a neighborhood of x if there exists ε > 0 such that Bε (x) ⊆ U . The following result is immediate. 1.3.2 Proposition. Let X be a metric space and take a point x ∈ X. (a) If ε > 0, then Bε (x) is a neighborhood of x. (b) If U is a neighborhood of x and U ⊆ V , then V is a neighborhood of x.
t u
1.3.3 Definition. Two metrics d and d0 in a set X are said to be equivalent if both determine the same neighborhoods of any point x ∈ X.
7
1.3 Neighborhoods and open sets
V
U
Bε (x)
x
Figure 1.2 A superset of a neighborhood of x is a neighborhood of x The following statement is easy to prove. 1.3.4 Proposition. Two metrics d and d0 in X are equivalent if and only if for each point x ∈ X the following holds: (a) Given ε > 0, there exists ε0 > 0 such that d0 (x, y) < ε0 =⇒ d(x, y) < ε, and (a) given δ 0 > 0, there exists δ > 0 such that d(x, y) < δ =⇒ d0 (x, y) < δ 0 . Proof: First notice that (a) and (b) are equivalent to the next, respectively: (a’) Given ε > 0, there exists ε0 > 0 such that Bε0 0 (x) ⊆ Bε (x), and (b’) given δ 0 > 0, there exists δ > 0 such that Bδ (x) ⊆ Bδ0 0 (x), where B stands for the balls with respect to the metric d and B 0 for the balls with respect to d0 . Hence (a) implies that the neighborhoods with respect to d are also neighborhoods with respect to d0 and (b) implies that the neighborhoods with respect to d0 are also neighborhoods with respect to d. Thus if (a) and (b) hold, then d and d0 are equivalent. Conversely, if d and d0 are equivalent and ε > 0, then by 1.3.2 Bε (x) is a neighborhood with respect to d0 , hence (a’) holds and thus (a) too. Analogously, Bε0 (x) is a neighborhood with respect to d and hence (b’) holds and thus (b) too. t u 1.3.5 Exercise. Take a metric d in a set X. Show that (a) the function d0 : X × X −→ R+ given by d0 (x, y) =
d(x, y) 1 + d(x, y)
defines a metric on X and this metric is equivalent to d, and
8
1 Metric spaces
(b) the function d00 : X × X −→ R+ given by d00 (x, y) = min{d(x, y), 1} defines another metric on X and this metric is also equivalent to d. From the previous exercise, we can conclude the following. 1.3.6 Proposition. Every metric space X with metric d has an equivalent metric d0 which is bounded. t u 1.3.7 Exercise. Given a metric d on X show that if k ∈ R, k > 0, then the function d0 given by d0 (x, y) = kd(X, y) defines an equivalent metric on X. We have also mentioned that the concept of open set in Euclidean spaces can be extended to metric spaces. 1.3.8 Definition. Let X be a metric space. We say that a subset A of X is open if A is a neighborhood of x for each x ∈ A. In other words, A ⊆ X is open if and only if for every x ∈ A there exists ε > 0 such that Bε (x) ⊆ A. By definition 1.3.3 we have that the open sets of a metric space depend only on the equivalence class of its metric. That is, equivalent metrics determine the same open sets. 1.3.9 Proposition. The following statements hold: (a) The open ball Bε (x) is an open set. (b) A subset A ⊂ X is open if and only if A is a union of open balls. Proof: Statement (a) follows from the triangle inequality and statement (b) follows from the definition of an open set. t u If we consider all open sets in a metric space X, we have the following. 1.3.10 Theorem. Let A be the set of all open sets in a metric space X. Then the following statements hold: (O1) Let I be an arbitrary set of indexes. If {Ai }i∈I is a family of elements in A, S then i∈I Ai is an element in A.
9
1.3 Neighborhoods and open sets
(O2) Let I be a finite set of indexes. If {Ai }i∈I is a family of elements in A, then T i∈I Ai is an element in A. S T In the particular case I = ∅, by definition, i∈∅ Ai = ∅ and i∈∅ Ai = X. Therefore (O1) and (O2) imply that ∅ and X belong to A. Proof: The sets X and ∅ are obviously open, that is, they belong to A. The former does because it is the universe where all balls are constructed, and the latter belongs to A by vacuity. Therefore, we may assume that I 6= ∅. (O1) It follows from 1.3.2. T (O2) Let I be finite and take x ∈ i∈I Ai . Since each Ai is open, there exists εi > 0 such that Bεi (x) ⊆ Ai , i ∈ I. Take ε = min{εi  i ∈ I}. Since I is finite, ε > 0 and clearly Bε (x) ⊆ Bεi (x) for all i ∈ I. Hence we have that T T t u Bε (x) ⊆ i∈I Ai and thus i∈I Ai is open (see Figure 1.3).
Figure 1.3 The intersection of finitely many open sets is open
1.3.11 Exercise. Take X = Rn and let d0 (x, y) =
n X i=1
be the metric of 1.2.5(a).
xi − yi 
10
1 Metric spaces
(a) Show that the neighborhoods of a point in Rn with the usual metric (Example 1.2.2(b)) are the same as those given by the metric d0 . (b) What can be said about the neighborhoods of a point in Rn defined using the metric of Example 1.2.2(e)? (c) Show that in general, if k = 1, 2, 3, . . . , the function v u n uX k dk (x, y) = t xi − yi k i=1
determines a metric and analyze the neighborhoods determined by it. (Hint: Use Minkowski’s inequality [2].) 1.3.12 Exercise. Take a discrete metric space X as in Example 1.2.2(d). Show that every point in X (seen as a onepoint set) is a neighborhood of itself. In other words, with respect to the discrete metric, every point is a neighborhood of itself, hence every point is an open set. 1.3.13 Exercise. Show that all metrics defined in 1.2.5(g) determine the same neighborhoods and prove that these neighborhoods are the same as those determined by the metric of 1.2.5(f). 1.3.14 Exercise. Show that the function d : I ×I −→ R given by d(s, t) = s−t2 satisfies (M1) and (M2). However, d does not satisfy (M3). A set X together with a function d : X × X −→ R which satisfies (M1) and (M2) is called a semimetric space. Thus I with d as defined above is a semimetric space. 1.3.15 Exercise. Let X be a metric space with metric d. A set C ⊆ X is said to be closed if its complement X − C is open. We say that a set B ⊆ X is bounded if B ⊂ Bn (x) for some point x ∈ X and some n ∈ N. Consider the set C(X) consisting of all nonempty bounded closed sets in X. For C ∈ C(X) and ε > 0 define the εneighborhood of C to be the set [ Nε (C) = {Bε (x)  x ∈ C} . Now define δ : C(X) × C(X) −→ R by δ(C, D) = inf{ε > 0  C ⊂ Nε (D) and D ⊂ Nε (C)} . Prove that δ is a metric that makes C(X) into a metric space. This metric is called the Hausdorff metric.
1.4 Convergence
11
1.3.16 Exercise. Let X be a metric space with metric d, and take a subset A ⊂ X . Show that A is bounded if and only if there is a positive number R such that d(x, y) ≤ R for every pair of elements x, y ∈ A. Define the diameter of a bounded set A to be the number diamA = sup{d(x, y)x, y ∈ A}. If ∆ = diamA exhibit a ball (giving its center and radius in terms of A and ∆) that contains A.
1.4
Convergence
Convergence of sequences is an important concept in analysis in Euclidean spaces. This concept can be defined in a similar way in any metric space. 1.4.1 Definition. Let (xn )n∈N be a sequence of points in a metric space X. We shall say that (xn ) converges to x, written xn → x, if for every neighborhood V of x there exists a number n0 ∈ N such that for every n ≥ n0 , xn ∈ V . If a sequence converges to x, we say that it is convergent. and we call x the limit of the sequence. 1.4.2 Exercise. Show that one obtains the same convergence concept if instead of asking V to be a neighborhood of x, we only require it to be a ball with center in x. The metric does not play the essential role in this concept of convergence. The concept lies rather on the neighborhoods and not explicitly on the metric. Since there are different metrics that determine the same neighborhoods, we have the same concept of convergence in these metric spaces. In other words, the concept of convergence depends exclusively upon the neighborhoods, or upon the open sets of the space, rather than upon the metric. We shall see below how we can axiomatize the concept of neighborhood without using the concept of metric. Therefore we shall have an associated concept of convergence of sequences. 1.4.3 Examples. In each of the next cases we shall refer to the metric spaces of the Examples 1.2.2 using the corresponding letter. (a) Convergence in R with the usual metric is the usual convergence. (b) Convergence in Rn with the usual metric is the usual convergence. (d) Convergent sequences in a discrete metric space are the almost constant sequences, that is, they are the sequences such that xn = x for all n > n0 (some n0 ∈ N).
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1 Metric spaces
(e) According to the comments about this metric made above, its associated neighborhoods are the usual neighborhoods in Rn , therefore, convergence in this metric is the usual convergence. (g) Here we have a concept of convergence of functions in terms of integrals. Consider a sequence of functions (fn ), where for each n, fn : I −→ Rn . The sequence is said to be pointwise convergent to a function f : I −→ Rn , if for each t ∈ I the sequence (fn (t)) of points in Rn converges in the usual way in Rn to the point f (t). Consider the following question: Does there exist a metric on the set X = {x : I −→ R  x is a continuous function} that induces pointwise convergence? The answer is no. We shall give below a more general concept of convergence that will depend rather on the neighborhoods and not on the metric. Dependance on the metric makes the concept much too restrictive. This is one of the several reasons why it is convenient to axiomatize the concept of open set in a metric space, to put aside all problems related to the metric.
1.5
Pseudometric spaces
Before passing to the axiomatization of the structure of the open sets in a metric space, it is convenient to introduce a generalization of the concept of metric space, which shares several of its properties and is frequently used as a source of examples. It has the same structure of its open sets. 1.5.1 Definition. Let X be a set and let d : X × X −→ R be a function that fulfills axiom (M3) of a metric and, instead of axiom (M1), it satisfies the axiom (SM1) d(x, x) = 0 ∀x ∈ X. Such a function is called a pseudometric in X and X together with d is called a pseudometric space. Notice that in a pseudometric space the “pseudodistance” between two points may be zero even though the two points are different. 1.5.2 Remark. As in Remark 1.2.3, one may show that if axioms (SM1) and (M3) hold, then d(x, y) ≥ 0 for all x, y ∈ X, and axiom (M2) holds.
13
1.5 Pseudometric spaces
1.5.3 Exercise. Show that for every X 6= ∅, the function d : X × X −→ R+ given by d(x, y) = 0 for any x, y ∈ X is a pseudometric. This is the socalled indiscrete pseudometric. In the same way as in the case of metric spaces, we can define open sets in pseudometric spaces. Let X be a pseudometric space with a pseudometric d and take a subset A ⊂ X. We say that A is open if for each point x ∈ A there exists ε > 0, such that the open pseudoball Bε (x) = {y ∈ X  d(x, y) < ε} with center x and radius ε lies in A. 1.5.4 Exercise. Show that the collection A of open sets in a pseudometric space X satisfies conditions (O1) and (O2) in 1.3.10. 1.5.5 Exercise. (a) Let X be a pseudometric space with pseudometric d and let ∼ be the relation given by x ∼ y ⇔ d(x, y) = 0. Show that ∼ is an equivalence relation in X. e = X/∼ consider the function In the set of equivalence classes X e ×X e → R+ , de : X e given by d([x], [y]) = d(x, y), where [x], [y] denote the corresponding equivae The metric space lence classes. Show that de is a welldefined metric on X. e with the metric de is called metric identification of X with respect to the X pseudometric d. e related to those of X? Give a characterization (b) How are the open sets of X of them. 1.5.6 Remark. Pseudometrics arise naturally in functional analysis. For instance, consider the space F(X) of realvalued functions f : X −→ R together with a special point x0 in the set X. This point then induces a pseudometric on the space of functions, given by d(f, g) = f (x0 ) − g(x0 ) for f, g ∈ F(X). 1.5.7 Exercise. (a) Define an equivalence relation in the square I × I in such a way, that it identifies points of the form (0, t) with those of the form (1, t). This relation
14
1 Metric spaces
corresponds to gluing the left edge of the square with the right edge to obtain a cylinder. Construct a pseudometric on I × I, whose metric identification provides the cylinder with a metric. (Hint: Let d be the usual metric on the plane R2 and for x, Y ∈ I × I, define d0 (x, y) = min{d(x, y), d(x + (1, 0), y)}. Check that d0 is a pseudometric with the desired property. The induced metric measures the distance between two points of the cylinder in a “geodesic” way, that is, “walking” over the surface of the cylinder.) (b) Extend the equivalence relation of (a) to include now the equivalence of a point of the form (s, 0) with the point (s, 1). This new relation corresponds to gluing the left edge of the square with the right edge, as well as the bottom edge with the top edge, to obtain a torus. Construct a pseudometric on I ×I, whose metric identification provides the torus with the ”geodesic” metric. (Hint: Let d be the usual metric on the plane R2 and for x, y ∈ I × I, define d0 (x, y) = min{d(x, y), d(x + (1, 0), y), d(x + (0, 1), y), d(x + (1, 1), y)}. Check that d0 is a pseudometric with the desired property. This is the “geodesic” metric on the torus.) (c) Define an equivalence relation in the square I × I in such a way, that it identifies points of the form (0, t) with those of the form (1, 1 − t). This relation corresponds to gluing the left edge of the square with the right edge, but in opposite directions, to obtain a Moebius band. Construct a pseudometric on I ×I, whose metric identification provides the Moebius band with a metric. (Hint: Let d be the usual metric on the plane R2 and for x = (s1 , s2 ), y = (t1 , t2 ) ∈ I × I, define d0 (x, y) = min{d(x, y), d(x0 , y), d(x, y 0 )}, where x0 = (s1 + 1, 1 − s2 ) and y 0 = (t1 + 1, 1 − t2 ).) 1.5.8 Exercise. (a) Prove that if X = {x : I −→ R  x is an integrable function }, then s Z 1 d(x, y) = (x(t) − y(t))2 dt 0
defines a pseudometric. (b) Describe the metric identification of the pseudometric space of (a). 1.5.9 Exercise. Notice that Definitions 1.3.1 and 1.3.8 make sense for pseudometric spaces and prove Proposition 1.3.10 in this more general case. 1.5.10 Exercise. Prove that for the indiscrete pseudometric (see 1.5.3) the only open sets are X and ∅. What is the metric identification in this case?
15
1.5 Pseudometric spaces
1.5.11 Exercise. Let X be a set, Y a metric space with a metric d0 and f : X −→ Y an arbitrary function. Show that the function d : X × X −→ R given by d(x, y) = d0 (f (x), f (y)) is a pseudometric. When is d a metric? Show that the statement is equally valid if d0 is only a pseudometric. 1.5.12 Exercise. Do the following functions define a metric on R2 ? Do they define a pseudometric? (a) d((x1 , x2 ), (y1 , y2 )) = y1 − x1 . (b) d((x1 , x2 ), (y1 , y2 )) = x1 x2 − y1 y2 . (c) d((x1 , x2 ), (y1 , y2 )) = x1 + y1 − x2 − y2 .
16
1 Metric spaces
Chapter 2
Topological spaces
In this chapter we shall give the basic definitions of topological space, open and closed sets, neighborhoods, and all definitions related with the description of a topological space. All these concepts are axiomatically defined taking as a model the corresponding concepts that one has in metric or pseudometric spaces. This way, metric and pseudometric spaces will be examples of topological spaces.
2.1
Basic definitions: open sets and neighborhoods
In Proposition 1.3.10 of the previous chapter, two properties of the family of open sets of a metric space were proved, namely (O1) and (O2). Further, in Exercise 1.5.4 we asked to prove the same properties for the open sets of a pseudometric space. They suggest what the basic definition of a topological space should be. This will be the starting point of the rest of this book. 2.1.1 Definition. Let X be a set. A topology in X is a family A of subsets of X, called the open sets, which satisfies the following two axioms: (O1) If {Ai }i∈I ⊆ A, where I is an arbitrary set, then (O2) If {Ai }i∈I ⊆ A, where I is a finite set, then
T i∈I
S i∈I
Ai ∈ A.
Ai ∈ A.
In particular, since X is the intersection of an empty family, X lies in A. Moreover, since ∅ is the union of an empty family, ∅ also lies in A. The pair (X, A) is called a topological space, which will be denoted simply by X if there is no danger of confusion with respect to the topological structure given by its open sets. The set X will be called the underlying set of the topological space. 2.1.2 Examples. The following are topologies. (a) The family A consisting of all subsets of a set X. This is called the discrete topology in X. The corresponding topological space is called discrete space. Therefore, in a discrete space X, every subset of X is open. 17
18
2 Topological spaces
(b) The family A consisting only of ∅ and X. This is called the indiscrete topology in X (this is also called the trivial topology. The corresponding topological space is called indiscrete space. Therefore, in an indiscrete space X, the only open sets are ∅ and X. (c) The family A of all open sets of a metric space X. This is called a metrizable topology. The corresponding topological space is usually called metrizable space. In particular, if X has the discrete metric, then A is the discrete topology, and X is a discrete space. Therefore, discrete spaces are metrizable. (d) The family A of all open sets of a pseudometric space X. This is called a pseudometrizable topology. The corresponding topological space is called pseudometrizable space. In particular, if X has the indiscrete pseudometric, then A is the indiscrete topology, and X is an indiscrete space. Therefore, indiscrete spaces are pseudometrizable. Notice that indiscrete spaces are not metrizable, unless they have at most one point. (e) The family A consisting of all subsets of a set X whose complement is finite and of ∅. If X es finite, then this topology is clearly the discrete topology. If X is infinite, then we call it the cofinite topology. (f) The family A consisting of all subsets of a set X whose complement is at most countable and of ∅. If X is countable, then this topology is clearly the discrete topology. If X es uncountable, then we call it the cocountable topology. 2.1.3 Exercise. Let X be a topological space and take a fixed subset S ⊂ X. Prove that the set {A ∪ (B ∩ S)  A, B are open sets in X} is another topology on the underlying set of X. By the discussion of the previous chapter, we know that different metrics on the same set X may determine the same open sets and therewith, also the same topology, as it was the case in Exercise 1.3.11, for instance. In two equivalent metrics, convergence of sequences is the same, i.e. a sequence is convergent in one metric if and only if it is convergent in the other. This suggests that convergence is a topological concept, rather than a metric concept. 2.1.4 Definition. Given a sequence (xn ) in a topological space, we say that it converges to a point x, if for every open set A containing x, the sequence lies inside A from certain element on. In symbols, xn → x if for each open set A such that x ∈ A, there exists n0 ∈ N such that xn ∈ A for all n ≥ n0 . We say that x is a limit of the sequence.
2.1 Definitions: open sets and neighborhoods
19
2.1.5 Exercise. Analyze and describe the convergence of sequences in all the topologies of the examples 2.1.2. As it is the case with metric spaces, in topological spaces we also can define the concept of neighborhood. 2.1.6 Definition. Let X be a topological space and take x ∈ X. We define a neighborhood of x to be a set U ⊆ X such that there exists an open set A in X that satisfies x ∈ A ⊆ U . This definition is clearly consistent with the one given for metric spaces 1.3.1, since an open ball in a metric space is an open set. Hence a neighborhood of a point in a metric space is a neighborhood of the same point in the topological space determined by the metric, and conversely. The definition of an open set in a metric space 1.3.8 becomes, for topological spaces, the following result. 2.1.7 Theorem. Let X be a topological space. A subset A ⊆ X is open if and only if A is a neighborhood of every x ∈ A. Proof: If A is an open set, then it is clearly a neighborhood of all its points. Conversely, assume that A is a neighborhood of each of its points, and take x ∈ A. Since A is a neighborhood of x, there exists an open set Ax such that S x ∈ Ax ⊆ A. Therefore, A = x∈A Ax , and by axiom (O1), A is open. t u From here on we shall denote by NxX the set of all neighborhoods of a point x in a topological space X. When there is no danger of confusion, we denote this set simply by Nx . The family N = {Nx  x ∈ X} will be called the neighborhood system for the topology of X. 2.1.8 Note. One may restate the definition of convergence given in Exercise 2.1.5 by saying that a sequence (xn ) in a topological space X converges to x, in symbols xn → x, if for every V ∈ Nx there exists n0 ∈ N such that xn ∈ V for all n ≥ n0 . 2.1.9 Proposition. The neighborhood system N for the topology of X has the following properties. (N1) V ∈ Nx , V ⊆ U ⇒ U ∈ Nx . (N2) Vi ∈ Nx , i ∈ I, I finite, ⇒
T i∈I
Vi ∈ Nx .
20
2 Topological spaces
(N3) V ∈ Nx ⇒ x ∈ V . (N4) U ∈ Nx ⇒ ∃ V ∈ Nx such that U ∈ Ny ∀ y ∈ V . Proof: (N1) It is clear. (N2) It follows from (O2). (N3) It is obvious by definition. (N4) Let V be an open set in X such that x ∈ V ⊆ U . Therefore V is a neighborhood of x and for each y ∈ V , y ∈ V ⊆ U . Thus U is a neighborhood of y. t u 2.1.10 Definition. Take a set X and let Nx , x ∈ X, be a family that fulfills conditions (N1)–(N4) of the previous proposition. Such a family N = {Nx  x ∈ X} is called a neighborhood system on X. A neighborhood system on a set X determines a topology on X as follows. 2.1.11 Theorem. Let X be a set and N a neighborhood system in X. Then there exists a unique topology on X that has N as its neighborhood system. Proof: First we prove that if such topology exists, then it is unique. Indeed, Theorem 2.1.7 states that a set A ⊆ X is open if and only if A is a neighborhood of each of its points. In other words, A is open if and only if A ∈ Nx for all x ∈ A. This fact characterizes the open sets uniquely. Hence the topology must be unique. This allows us now to define the topology. Indeed, define A = {A ⊆ X  A ∈ Nx ∀x ∈ A} . We now have to prove that A is in fact a topology. We must check that the axioms hold. First notice that ∅ ∈ A by vacuity, and X ∈ A by Axiom (N1). S (O1) Let {Ai }i∈I ⊆ A be a nonempty family of sets in A. If x ∈ i∈I Ai , then S x ∈ Ai for some i ∈ I. Since Ai ∈ A, Ai ∈ Nx . By (N1), i∈I Ai ∈ Nx and S therefore i∈I Ai ∈ A.
2.1 Definitions: open sets and neighborhoods
21
T (O2) Let {Ai }i∈I ⊆ A be a finite nonempty family of sets in A. If x ∈ i∈I Ai , then x ∈ Ai for all i ∈ I. Since Ai ∈ A, Ai ∈ Nx for all i ∈ I. By (N2), T T i∈I Ai ∈ Nx and therefore i∈I Ai ∈ A. We have shown that A is a topology on X. We now have to study the neighborhoods in this topology. We shall prove that they are precisely the elements of the given neighborhood system. Namely, let V be a neighborhood of x with respect to the topology A, that is, there exists a set A ∈ A such that x ∈ A ⊆ V . By definition of A, one has A ∈ Nx , and by (N1), V ∈ Nx . Conversely, assume that U ∈ Nx . Define A = {y ∈ X  U ∈ Ny }. In particular, x ∈ A and, by (N3), y ∈ U for all y ∈ A. Thus x ∈ A ⊆ U . Hence it is enough to check that A ∈ A. Take y ∈ A. Since U ∈ Nx , then by definition of A and by (N4) we know that there exists V ∈ Ny such that U ∈ Nz for all z ∈ V . Hence V ⊆ A and by (N1), A is a neighborhood of y. Therefore A ∈ A as desired. t u The construction given in the proof of 2.1.11 of the open sets A leads to the next. 2.1.12 Definition. Let X be a topological space and take a subset A ⊆ X. Define the interior of A as the set A◦ = {x ∈ A  A ∈ Nx } . A point x ∈ A◦ is called an interior point of A. 2.1.13 Theorem. Let X be a topological space and A be a subset of X. Then [ A◦ = {B ⊆ A  B is open in X}. Therefore A◦ is an open set. Indeed, A◦ is the largest open set of X contained in A. Proof: Take x ∈ A◦ . Then A is a neighborhood of x. Hence there is an open set S B ⊂ X such that x ∈ B ⊆ A. Thus, clearly, A◦ ⊆ {B ⊆ A  B is open in X}. S Conversely, take x ∈ {B ⊆ A  B is open in X}. Then x ∈ B ⊆ A for S some open set B ⊂ X. Therefore A ∈ Nx , that is, x ∈ A◦ . Hence, {B ⊆ A  B is open in X} ⊆ A◦ . t u 2.1.14 Corollary. A is open in X if and only if A = A◦ .
t u
This corollary is an equivalent statement of Theorem 2.1.7, which was proved above.
22
2 Topological spaces
2.1.15 Definition. Let X be a topological space and A be a subset of X. The assignment A 7→ A◦ is called interior operator of the topology of X. It is easy to prove the following. 2.1.16 Theorem. Let X be a topological space. The interior operator has the following properties: (I1) X ◦ = X. (I2) A◦ ⊆ A. (I3) (A◦ )◦ = A◦ . (I4) (A ∩ B)◦ = A◦ ∩ B ◦ . (I5) A ⊆ B ⇒ A◦ ⊆ B ◦ . (I6) (A ∪ B)◦ ⊇ A◦ ∪ B ◦ .
t u
2.1.17 Exercise. (a) Prove the previous theorem. (b) Show with an example that, in general, the equality in (I6) does not hold. (c) Prove that (I5) and (I6) are consequences of (I4). 2.1.18 Definition. Let X be a set. An operator in the subsets of X, A 7→ A◦ , that satisfies (I1)–(I4), is called interior operator on the set X. In the same spirit of Theorem 2.1.11, that characterizes the topology of a topological space in terms of its neighborhoods, that is, in terms of the axioms (N1)–(N4), we have the following. 2.1.19 Theorem. Let X be a set and let A 7→ A◦ be an interior operator on X. Then there exists a unique topology A on X, whose interior operator is the given one. Proof: Define A = {A ⊆ X  A = A◦ }. Then A is the desired topology on X.
t u
2.1.20 Exercise. Prove that indeed A, as defined in the previous proof, is a topology. Furthermore, prove that it is the only topology on X that has A 7→ A◦ as its interior operator.
23
2.2 Closed sets
2.1.21 Exercise. Let X be a set and and take a fixed subset A0 ⊆ X. Define an operator A 7→ A◦ by ½ X if A = X , A◦ = A − A0 if A 6= X , where A − A0 = {x ∈ A  x ∈ / A0 }. Prove that this is an interior operator on X. Analyze this operator in the cases A0 = X or A0 = ∅ and explain what topology is described in each case. 2.1.22 Exercise. Describe the interior operator for the cofinite, resp. the cocountable, topology on an infinite, resp. uncountable, set X.
2.2
Closed sets
The subsets of a topological space X that we shall call “closed” determine the topology, just as the open sets do. However, their properties are quite different to those of the open sets. Therefore, it is convenient to study them separately. 2.2.1 Definition. A point x in a topological space X is called limit point of a set A ⊆ X if every neighborhood V of x in X meets A, i.e. is such that V ∩ X 6= ∅. Let A = {x ∈ X  x is a limit point of A} . The set A is called closure a A. It is easy to prove have the following. 2.2.2 Proposition. The following two conditions hold and are equivalent: (a) A = X − (X − A)◦ (b) X − A = (X − A)◦ .
t u
2.2.3 Remark. From the previous result we have that, if X − A is open, then X − A = (X − A)◦ = X − A and hence A = A. 2.2.4 Definition. A subset A of a topological space X is said to be closed if X − A is open.
24
2 Topological spaces
Thus a subset A of a topological space X is closed if and only if A = A. In particular, we have that X and ∅ simultaneously open and closed. From the axioms (O1) and (O2) for the open sets of a topological space one has the following. 2.2.5 Theorem. Let X be a topological space. Then the family C of the closed sets in X satisfies the following axioms: (C1) Bi ∈ C, i ∈ J , ⇒
T i∈J
Bi ∈ C.
(C2) Bi ∈ C, i ∈ J , J finite, ⇒
S i∈J
Bi ∈ C.
t u
2.2.6 Exercise. Prove the previous theorem with all details. 2.2.7 Examples. 1. In a discrete space all subsets are open and closed, while in an indiscrete space, the only closed subsets are X and ∅. 2. In the real line R with the usual topology one has, for instance, that the singular sets {t} ⊂ R, as well as [a, b], (−∞, b], and [a, +∞) are closed subsets. On the other hand, an open interval (a, b), a < b, is open and its closure is the closed interval [a, b]. However, [ {t} (a, b) = a 0} ,
si x 6= 0 ,
y
B0 = {(−ε, ε) ∪ (−∞, −n) ∪ (n, ∞)  ε > 0, n ∈ N} .
32
2 Topological spaces
(a) Prove that Bx is a neighborhood basis for each x ∈ R. (b) Describe the closed sets and the closure operator in the topology determined by these neighborhood bases. The space L obtained from R with this topology is called wrapped line.
2.5
Continuity
2.5.1 Definition. Let X and Y be metric spaces with metrics d and d0 , respectively. A function f : X −→ Y is said to be continuous at a point x ∈ X if given ε > 0 there exists δ > 0such that, if d(y, x) < δ, then d0 (f (y), f (x)) < ε. One says that f is continuous if it is continuous at each point x ∈ X.
The concept of continuity in topological spaces is, perhaps, the central concept of topology. Before introducing the general concept of continuity in topological spaces, it is convenient to state the concept in the case of metric spaces, which is an immediate generalization of the continuity in Euclidean spaces studied in elementary calculus courses. This means, in common language, that a continuous map preserves the “nearness”, in other words, it maps close points to close points.
Convention. In what follows, if it is not specified otherwise, the word map will mean a continuous function. 2.5.2 Examples. (a) Any constant map κy0 : X −→ Y , κy0 (x) = y0 , is clearly continuous. (b) The identity map idX : X −→ X is continuous. (c) The map f : R −→ R given by ½ 0 f (t) = 1
if t < 0 if t ≥ 0
is not continuous at t = 0. Clearly f does not map points close to 0 to points close to f (0) = 1. Namely, no matter how close t < 0 is to 0, we have d(f (t), f (0)) = d(0, 1) = 0 − 1 = 1.
33
2.5 Continuity
x
f (x)
V
U f
X
Y
Figure 2.3 A continuous map sends neighborhoods into given neighborhoods Given a map f : X −→ Y between metric spaces and a point x ∈ X, we can rewrite the definition of continuity at x as follows. 2.5.3 Proposition. A map f : X −→ Y is continuous at x if and only if, given ε > 0, there exists δ > 0 such that f (Bδ (x)) ⊂ Bε (f (x)). Equivalently, the map f is continuous at x if and only if, given ε > 0, there exists δ > 0 such that Bδ (x) ⊂ f −1 (Bε (f (x))). t u The last part of this assertion states, in other words, that f is continuous if and only if, for every ε > 0, f −1 (Bε (f (x))) is a neighborhood of x. This suggests how to formulate the general definition of a continuous map in arbitrary topological spaces. 2.5.4 Definition. Let X and Y be topological spaces and take f : X −→ Y . We say that f is continuous at a point x ∈ X, if, given a neighborhood V of f (x) in Y , the inverse image f −1 (V ) is a neighborhood of x in X. f is said to be continuous if it is continuous at any point in X. 2.5.5 Remark. For metrizable topological spaces, this definition of continuity is equivalent to the definition of continuity in metric spaces, as shown in Proposition 2.5.3. We also have the following characterization. 2.5.6 Proposition. A map f : X −→ Y is continuous at x ∈ X if and only if, given a neighborhood V of f (x), there exists a neighborhood U of x such that f (U ) ⊂ V . t u
34
2 Topological spaces
2.5.7 Theorem. Let X, Y , and Z be topological spaces and let f : X −→ Y be continuous at x and g : Y −→ Z be continuous at f (x). Then the composite g ◦ f : X −→ Z is continuous at x. Consequently, if f and g are continuous, then the composite g ◦ f is continuous. Proof: Take a neighborhood W of gf (x) = g(f (x)) in Z. Since g is continuous at f (x), the inverse image g −1 (W ) is a neighborhood of f (x) in Y . Analogously, since f is continuous at x, the inverse image f −1 (g −1 (W )) = (g ◦ f )−1 (W ) is a neighborhood of x in X. t u As we have seen, the topology of a space can be equivalently defined in several ways. Either giving its open sets, its closed sets, its interior operator, or its closure operator among others. Similarly, continuity can be characterized using any of these concepts. We have the following. 2.5.8 Theorem. Let X and Y be topological spaces and take a map f : X −→ Y . The following are equivalent: (a) f is continuous. (b) For any open set B ⊆ Y , the inverse image f −1 (B) ⊆ X is an open set. (c) For any set B ⊆ Y , f −1 (B ◦ ) ⊆ f −1 (B)◦ . (d) For any set A ⊆ X, f (A) ⊆ f (A). (e) For any closed set B ⊆ Y , the inverse image f −1 (B) ⊆ X is a closed set. Proof: (a)⇒(b) Let B be open in Y and take x ∈ f −1 (B). Thus f (x) ∈ B and since B is open in Y , B is a neighborhood of f (x). Since f is continuous (at x), f −1 (B) is a neighborhood of x. Hence, since x is arbitrary, f −1 (B) is open. (b)⇒(c) Take a set B ⊆ Y . By (b), f −1 (B ◦ ) is open and is a subset of f −1 (B). Therefore, it is also a subset of f −1 (B)◦ . (c)⇒(d) Take a set A ⊂ X. By (c), f −1 ((Y − f (A))◦ ) ⊂ f −1 (Y − f (A))◦ . Hence X − f −1 f (A) = f −1 (Y − f (A)) = f −1 ((Y − f (A))◦ ) ⊂ f −1 (Y − f (A))◦ = (X − f −1 f (A))◦ = X − f −1 f (A). Therefore, since A ⊂ f −1 f (A), one has A ⊂ f −1 f (A) ⊂ f −1 f (A). Taking the image under f of the first and the last term of the previous series we obtain f (A) ⊂ f (f −1 f (A)) ⊂ f (A). (d)⇒(e) Let B be closed in Y . By (d) and since f f −1 (B) ⊂ B, we have f (f −1 (B)) ⊂ f f −1 (B) ⊂ B = B. If we take inverse images, then we obtain
2.6 Homeomorphisms
35
f −1 f (f −1 (B)) ⊂ f −1 (B). But f −1 (B) ⊂ f −1 f (f −1 (B)), thus f −1 (B) ⊂ f −1 (B) and consequently f −1 (B) = f −1 (B) is closed. (e)⇒(a) Take any point x ∈ X and let V be a neighborhood of f (x) in Y . Without loss of generality, we may assume that V is open. Hence Y − V is closed and, by (e), we have that f −1 (Y − V ) = X − f −1 (V ) is also closed. Therefore, f −1 (V ) is open and since it clearly contains x, it is a (open) neighborhood of x in X. Thus f is continuous at x and since x is arbitrary, f is continuous. t u 2.5.9 Exercise. Let X be a topological space. Prove that the following are equivalent: (a) X has the indiscrete topology. (b) Given any topological space Y and any map f : Y −→ X, f is continuous. 2.5.10 Exercise. Let X be a topological space. Prove that the following are equivalent: (a) X has the discrete topology. (b) Given any topological space Y and any map f : X −→ Y , f is continuous. 2.5.11 Exercise. Let X and Y be topological spaces, each with with the cofinite topology (see 2.1.2(e)) and take f : X −→ Y . Give a necessary and sufficient condition (in terms of the finite subsets of X and Y ) in order for f to be continuous. 2.5.12 Exercise. Using only definitions and results from this and the previous chapter, show that the two maps R2 −→ R that map (s, t) to s + t and to st are continuous.
2.6
Homeomorphisms
A topological space is a set together with certain additional structure given by its open sets (or, equivalently, by its closed sets, its interior operator, or its closure operator). In algebra, for instance, the objets of study are also sets with another “algebraic” structure, which might be a group structure, a ring structure, etc. From the point of view of algebra, one does not distinguish between two objects, when they are isomorphic, that is, when there is a bijection between the two underlying sets which preserves the structure. Similarly, in the case of topological spaces one can speak of “isomorphic” spaces when there exists a bijection between their underlying sets which preserves the “topological structure”.
36
2 Topological spaces
2.6.1 Definition. Let X and Y be topological spaces and let f : X −→ Y be a map. We say that f is a homeomorphism if it is a bijective continuous map and its inverse map f −1 : Y −→ X is also continuous. If such a homeomorphism exists, then we say that the spaces X and Y are homeomorphic, and we denote this fact ≈ by f : X −→ Y or, more simply, by X ≈ Y (some authors write X ∼ = Y ). As we already said, a homeomorphism is a bijection which preserves the topological structure. This we shall see now. 2.6.2 Theorem. Let X and Y be topological spaces and f : X −→ Y a map. The following statements are equivalent: (a) f is a homeomorphism. (b) f is bijective and f (A) is open in Y if and only if A is open in X. (c) f is bijective and f (A) is closed in Y if and only if A is closed in X. Proof: It is enough to observe that for every subset A of X, the inverse image f −1 (f (A)) = A. With this fact the proof follows immediately from Theorem 2.5.8. t u Statement (b) above says that f not only establishes a bijection between the points of X and those of Y , but also between the open sets of X and the open sets of Y . Similarly, statement (c) says the same, but for the closed sets instead. It is in this sense that a homeomorphism establishes an equivalence not only between the points of the underlying sets, but also between their topological structures. 2.6.3 Remark. If f is continuous and bijective, then it is not necessarily a homeomorphism. For example, if X a discrete space with more than one element and Y is an indiscrete space with the same underlying set of X, then the identity map X −→ Y is bijective and continuous, but it is obviously not a homeomorphism. 2.6.4 Exercise. Prove that the map f : [0, 1) −→ S1 , given by f (t) = e2πit , is bijective and continuous, but it is not a homeomorphism. (Hint: The inverse map log : S1 → [0, 1) is not continuous at 1 ∈ S1 , since the inverse image of an open neighborhood of 0 ∈ [0, 1) of the form [0, ε) with ε < 1, is not open in S1 .) 2.6.5 Examples. (i) The identity of of a topological space X, idX , is a homeomorphism.
37
2.6 Homeomorphisms
(ii) If f : X −→ Y and g : Y −→ Z are homeomorphisms, then g ◦ f : X −→ Z is a homeomorphism. (iii) Consider the pierced unit sphere X = Sn − {N } ⊂ Rn+1 , where N = (0, . . . , 0, 1) is the north pole. The map p : X −→ Rn
(2.6.6) given by
µ p(x) =
x1 xn ,..., 1 − xn+1 1 − xn+1
¶ ,
where x = (x1 , x2 , . . . , xn+1 ) ∈ X, is a homeomorphism whose inverse is given by µ ¶ 2y1 2yn y2 − 1 −1 , p (y) = ,..., 2 , y2 + 1 y + 1 y2 + 1 where y = (y1 , . . . , yn ) ∈ Rn . The map p is the socalled stereographic projection. 2.6.7 Remark. The relation between topological spaces given by a homeomorphism is clearly an equivalence relation. 2.6.8 Examples. (i) Every affine, nonsingular transformation f : Rn −→ Rn , which is given by f (x) = x0 + L(x), where L is a linear isomorphism, is a homeomorphism. (ii) I n es homeomorphic to Bn . 2.6.9 Exercise. Show an explicit homeomorphism for Example 2.6.8(ii). 2.6.10 Exercise. Take X = {x, y}. Prove that A = {X, ∅, {x}} is a topology. The corresponding topological space is called Sierpi´ nski space. 2.6.11 Exercise. Write explicitly all possible topologies in a threeelement set X = {x, y, z}. Prove that among all of them, there are exactly nine such that the corresponding topological spaces are not homeomorphic. Determine which of them are comparable and in that case indicate which is finer. 2.6.12 Exercise. Show that in a metrizable space X the following statements hold:
38
2 Topological spaces
(a) Each onepoint set is the intersection of all its (open) neighborhoods. (b) Each onepoint set is a closed set. In Rn no point is open. Hence (a) implies that the intersection of open sets is not always an open set. 2.6.13 Exercise. Let X be a discrete space and A ⊂ X. Describe the boundary of A. 2.6.14 Exercise. Let X be a topological space and take A ⊂ X. Show that the boundary of A, ∂A, is a closed set. 2.6.15 Exercise. Let X be a metric space and take y ∈ X. Show that the function f : X −→ R given by f (x) = d(x, y) is continuous. 2.6.16 Exercise. A set A ⊂ Rn is called convex if for any two points x0 , x1 ∈ A, the line segment [x0 , x1 ] = {(1 − t)x0 + tx1  t ∈ I} is contained in A. Show that if A ⊂ Rn is a closed, bounded convex set, whose interior is nonempty, then A is homeomorphic to the nball Bn .
Chapter 3
Comparison of topologies
We have already seen that a given a set can be furnished with different topologies. In this chapter we shall discuss the relationship among the various topologies on the same underlying set. We shall introduce the concepts of topologies that are coarser or finer than other topologies. We shall as well answer the question of finding the coarsest and the finest topologies of a given family of topologies. This will be used to introduce the concepts of basis and subbasis of a topology, which are families of open sets smaller than the topology, but which however determine the topology.
3.1
Comparison of topologies
If A1 and A2 are two topologies on a set X, it is possible that the open sets according to one of the topologies, say A2 , are also open sets according to the other topology A1 , that is, A2 ⊂ A1 , or it might happen that the topologies cannot be compared. We have the following. 3.1.1 Definition. We say that a topology A2 is (strictly) coarser than a topology A1 , or that A1 is (strictly) finer than A2 , if A2 ⊂ A1 (and they are different). We also say that A1 and A2 are comparable, if either A2 ⊂ A2 or A2 ⊂ A1 . The motivation of this terminology has to do in some sense with the “size” of the open sets. The more open sets we have in X, the finer they must be in order to fit into X. 3.1.2 Examples. 1. The discrete topology on X is finer than any other topology on X. On the other hand, the indiscrete topology on X is coarser than any other topology on X. Moreover, if X has more than one point, then the discrete topology is strictly finer than the indiscrete topology on X, and correspondingly, the indiscrete topology is strictly coarser than the discrete topology on X. 39
40
3 Comparison of topologies
2. If X = {x, y}, then we can consider the topologies A1 = {X, ∅, {x}} and A2 = {X, ∅, {y}}. The topologies A1 and A2 are not comparable. The following result is clear from the definition. 3.1.3 Theorem. Call X1 a set X furnished with a topology A1 and call X2 the same set X with a topology A2 . Then id : X1 −→ X2 is continuous if and only if A1 is finer than A2 . t u 3.1.4 Theorem. Let A1 and A2 be topologies on X. Then the following statements are equivalent: (a) A1 is finer than A2 , i.e. A2 ⊂ A1 . (b) For each x ∈ X, any neighborhood of x according to A2 is also a neighborhood according to A1 , i.e. Nx2 ⊂ Nx1 . (c) Every closed set according to A2 is closed according to A1 , i.e. C2 ⊂ C1 . (d) For any set A ⊆ X, the closure of A according to A1 is contained in the 1 2 closure of A according to A2 , i.e. A ⊂ A . (e) For any set A ⊆ X, the interior of A according to A1 contains the interior of A according to A2 , i.e. A◦2 ⊂ A◦1 . Proof: By 3.1.3, statement (a) is equivalent to say that id : X1 −→ X2 is continuous. Thus, the equivalence of statements (b), (c), (d) and (e) with (a) is an t u immediate consequence of Definition 2.5.4 and of Theorem 2.5.8. 3.1.5 Remark. Two given topologies on a set might not be comparable. A simple example is the set X = {x, y} with the topologies A1 = {X, ∅, {x}} and A2 = {X, ∅, {y}}. On the other hand, the relation “finer than” is transitive, namely, if A1 is finer than A2 and A2 is finer that A3 , then A1 is finer than A3 (i.e. A1 ⊃ A2 , A2 ⊃ A3 ⇒ A1 ⊃ A3 ). One also has that if A1 is finer than A2 and A2 is finer than A1 , then A1 = A2 , (i.e. A1 ⊃ A2 , A2 ⊃ A1 ⇒ A1 = A2 ). In other words we can say that the family of all topologies on a given set constitute a partially ordered set. See 7.3.1 below. Given any two topologies on a set X, there is always one which is finer than each of the two, namely the discrete topology. There is another topology which is coarser than both, namely the indiscrete topology. One may ask the following question: Is there among all topologies in X finer than A1 and A2 a coarsest
41
3.2 Intersection of topologies
topology? Another question: Is there among all topologies in X coarser than A1 and A2 one which is the finest topology? In what follows we shall study these questions. 3.1.6 Exercise. Consider the set X = {x, y, z} and take A1 = {∅, {x}, {x, y}, X} and A2 = {∅, {x}, {y, z}, X}. Show that A1 and A2 are topologies. Find the coarsest topology that contains both of them. Find the finest topology contained in both of them.
3.2
Intersection of topologies
Assume given a a family {Aλ  λ ∈ Λ} of topologies on a set X. Our goal is to find a topology A which is maximal with the property that its open sets are also open sets of Aλ for all λ. In other words, we wish to construct the finest topology which is coarser than all topologies Aλ . The ideal candidate for that is the intersection of all members Aλ of the given family. 3.2.1 Definition. Let {Aλ  λ ∈ Λ} be a family of topologies on the same set X. We define the infimum of the family, denoted by inf{Aλ  λ ∈ Λ}, to be the maximal (finest) topology A among all topologies which are coarser than all topologies Aλ . We have the following. 3.2.2 Proposition. Let {Aλ  λ ∈ Λ} be a family of topologies on the same set X. Then \ A= Aλ λ∈Λ
is a topology. Clearly it is the finest topology among all topologies that are coarser than Aλ . Hence \ inf{Aλ  λ ∈ Λ} = Aλ . λ∈Λ
t u The following result provides us with a characterization of the infimum of a family of topologies. It has the form of a universal property. 3.2.3 Theorem. Given a family {Aλ  λ ∈ Λ} of topologies on a set X, its infimum A is characterized by the following universal property, given in two parts. Denote by Xλ the set X with the topology Aλ and by X the set X with the topology A. Then
42
3 Comparison of topologies
(a) id : Xλ −→ X is continuous for all λ. (b) Given a function of sets f : X −→ Y such that fλ = f : Xλ −→ Y is continuous for all λ ∈ Λ, then f : X −→ Y is continuous. Put in a diagram Xλ id
fλ
 f ² 
/Y >
X
f is continuous ⇔ fλ is continuous ∀ λ ∈ Λ. Proof: (a) Since every open set A of X is open in Xλ , clearly id : Xλ −→ X is continuous for all λ ∈ Λ. (b) If f : Xλ −→ Y is continuous for all λ ∈ Λ, then for each open set B in Y , is in Aλ for all λ. Consequently f −1 (B) lies in A, and thus f : X −→ Y is continuous. f −1 (B)
Conversely, if the topology A has the property of the statement, and since id : Xλ −→ X is continuous, A lies in each Aλ . Therefore A is contained in T 0 λ∈Λ Aλ . Let now X be the set X with the topology of the intersection and take Y = X 0 and f = id, which clearly satisfy the property. Consequently id : X −→ X 0 T is continuous and hence every open set of λ∈Λ Aλ is an open set in A. Thus T t u A = λ∈Λ Aλ . 3.2.4 Exercise. Consider a set X and take B ⊆ X. (a) Show that A = {X, ∅, B} is a topology on X. In particular, {{a, b, c}, ∅, {b}} is a topology on the set {a, b, c}. (b) Let f : X −→ {a, b, c} be a function and assume B = f −1 {b}. Show that the topology A defined in (a) is the infimum of all topologies on X that make f continuous. (c) Describe the interior operator and the closure operator for the topology A of (a).
3.3
Supremum of a family of topologies
As before, assume given a a family {Aλ  λ ∈ Λ} of topologies on the set X. Our question now is about the existence of a topology A which is minimal with the
3.3 Supremum of a family of topologies
43
property that it contains all open sets of all the topologies Aλ . In other words, the topology A must be the coarsest of all topologies that are finer than all the topologies of the family. The first that comes into mind is to consider the union S λ∈Λ Aλ , but of course we should ask us if this union is a topologie. Clearly, it is not. 3.3.1 Definition. Let {Aλ  λ ∈ Λ} be a family of topologies on the same set X. We define the supremum of the family, denoted by sup{Aλ  λ ∈ Λ}, to be the minimal (coarsest) topology A among all topologies which are finer than all topologies Aλ . 3.3.2 Example. Take X = {x, y, z} and the topologies A1 = {X, ∅, {x}} and A2 = {X, ∅, {y}} on X. Clearly the union A1 ∪ A2 is not a topology. However the only set missing in order to have a topology is {x, y}, namely the union of the only two nontrivial open sets in A1 and A2 . Thus A1 ∪ A2 ∪ {{x, y}} is the supremum of A1 and A2 . Before we take in our hands the solution of the concrete problem we are facing, we shall analyze another more general problem. Let F be an arbitrary family of subsets of a set X. If A is a topology on X such that all elements of F are open sets of A, then not only F ⊂ A, but A contains all finite intersections of elements of F. Let I denote the family of all finite intersections of elements of F. Hence we have F ⊂ I ⊂ A. In particular, since X is the intersection of an “empty family”, we have X ∈ I. The family I is not yet a topology; the desired topology A must also contain all unions of elements in I. So take now the family U of all unions of elements of I. In particular, ∅ ∈ U, since ∅ is the union of an “empty family.” Thus we have F ⊂ I ⊂ U ⊂ A. 3.3.3 Proposition. The family U is a topology on X. Clearly U is the coarsest topology for which the elements F are open sets. t u 3.3.4 Definition. The family U is called the topology generated by F. The original family F is called subbasis of the topology U. The elements of F are called subbasic open sets. Thus a subbasis F for a topology U is a family of open sets in U such that any open set in U is a union of finite intersections of open sets in F.
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3 Comparison of topologies
Every topology U has at least one subbasis, namely U itself. This concept of subbasis is quite useful. An example of this is the following. 3.3.5 Theorem. Let X be topological space and F a subbasis of its topology. Take a map f : Y −→ X. Then f is continuous if and only if f −1 (S) is an open set for all S ∈ F. t u Now we may come back to the starting problem of this section. Given the S family {Aλ  λ ∈ Λ}, take F = λ∈Λ Aλ . Thus the supremum of the family {Aλ } is the topology U determined by F, namely, we have the following. 3.3.6 Theorem. The supremum of a family {Aλ  λ ∈ Λ} of topologies on a set X consists of arbitrary unions of finite intersections of open sets in Aλ for any λ. t u 3.3.7 Theorem. The supremum of a family {Aλ  λ ∈ Λ} of topologies on a set X is characterized by the following universal property, given in two parts. Denote by Xλ the set X with the topology Aλ and by X the set X with the topology A. Then (a) id : X −→ Xλ is continuous for all λ. (b) Given a function of sets f : Y −→ X such that fλ = f : Y −→ Xλ is continuous for all λ ∈ Λ, then f : Y −→ X is continuous. Put in a diagram
f
Y



fλ
X > ²
id
/ Xλ
f is continuous ⇔ fλ is continuous ∀ λ ∈ Λ. t u 3.3.8 Exercise. Using 3.3.5, show in the same spirit as 3.2.3, the previous theorem. 3.3.9 Exercise. Show that the open half lines (−∞, b) y
(a, +∞) ,
constitute a subbasis of the usual topology of R.
a, b ∈ R ,
45
3.4 Basis of a topology
3.3.10 Exercise. Let X be an (infinite) set. Show that the family S = {X −{x}  x ∈ X} is a subbasis for the cofinite topology on X (2.1.2(e)). 3.3.11 Exercise. Let X be a set and let Kλ be a topological space, λ ∈ Λ. Consider a family of functions {fλ : Kλ −→ X  λ ∈ Λ} and put Aλ = {A ⊆ X  fλ−1 A ⊆ Kλ is open}. (a) Show that Aλ is a topology on X. Call Xλ the set X furnished with this topology. (b) Let A be the infimum of the family of topologies {Aλ  λ ∈ Λ}. Show that A is the supremum of all topologies on X that make fλ : Kλ −→ X continuous for all λ ∈ Λ. (c) Let g : X −→ Y be a function, where X denotes the ste X furnished with the topology A of (b), and let Y be an arbitrary topological space. Show that g is continuous if and only if g ◦ fλ : Kλ −→ Y is continuous for all λ ∈ Λ. (d) For any two elements λ, µ ∈ Λ let ϕλµ : Kλ −→ Kµ be a continuous map such that the equality fµ ◦ ϕλµ = ϕλ holds. Assume that the maps ϕλµ have the following two properties: (i) Given λ ∈ Λ, the map ϕλλ : Kλ −→ Kλ is the identity. (ii) Given λ, µ, ν ∈ Λ, one has the equality ϕλν = ϕµν ◦ ϕλµ : Kλ −→ Kµ −→ Kν . For each λ ∈ Λ, take a function gλ : Kλ −→ Y such that gµ ◦ fµλ = gλ , and let g : X −→ Y satisfy g ◦ ϕλ = gλ . Show that g is continuous if and only if gλ is continuous for each λ ∈ Λ. In a diagram Kλ B
BB BB gλ BBB Ã
fλ
Y
/X ~ ~ ~~ ~~ g ~ ~Ä
g is continuous ⇔ gλ is continuous for all λ ∈ Λ .
3.4
Basis of a topology
In a metric or pseudometric space, balls are the building blocks that form all open sets, namely, any open set is a union of balls. As we already saw in the previous section, in general the open sets of a topological space are unions of finite
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3 Comparison and generation of topologies
intersections of open sets in a subbasis of the topology. This brings us to the following. 3.4.1 Definition. A family B of open sets in a topological space X is called a basis of (the topology of) X if every open set in X is a union of elements of the family B. The elements of B are called basic open sets. In particular, every basis of a topology is a subbasis. We shall use below the following notion. 3.4.2 Definition. Let M be a set and ≤ a relation on M . If the pair (M, ≤) satisfies the following axioms (OR1) a ≤ a (the relation is reflexive). (OR2) a ≤ b, b ≤ c =⇒ a ≤ c (the relation is transitive). (OR3) a ≤ b, b ≤ a =⇒ a = b (the relation is antisymmetric). (OR4) For all a, b ∈ M , either a ≤ b or b ≤ a. then we say that the relation ≤ is a total order in M and that in that case, M is a totally ordered set. If a ≤ b and a 6= b, then we write a < b. 3.4.3 Examples. (a) If X is a metric or pseudometric spave, then both B = {Bε (x)  ε > 0, x ∈ X} and B0 = {B1/n (x)  n ∈ N, x ∈ X} are bases for the topology of X. (b) If S is a subbasis for a topology on X, then B = {S1 ∩ · · · ∩ Sk  S1 , . . . , Sk ∈ S} is a basis for that topology. (c) In R, the open intervals constitute a basis for the usual topology.
47
3.4 Basis of a topology
(d) If X is a totally ordered set with order ≤ (see 3.4.2), then the open intervals (a, b) = {x  a ≤ x ≤ b, a 6= x 6= b}, a, b ∈ X, together with the open rays (−∞, b) = {x ∈ x < b} and (a, ∞) = {x ∈ X  a < x}, constitute a basis for a topology, called order topology. This topology has as a subbasis the open rays (−∞, b) and (a, ∞), for all a, b ∈ X. Since by (c), the usual topology of R has the open intervals as basis, this topology is an order topology associated to the usual order of the reals. 3.4.4 Exercise. Consider the set R2 and define on it the lexicographic order, namely the order given by (x, y) ≤ (x0 , y 0 ) if either x ≤ x0
or if x = x0
and y ≤ y 0 .
Show that this is a total order. Furthermore, describe the open intervals and the open rays, and describe the corresponding order topology. 3.4.5 Exercise. The natural numbers N are an ordered set with minimal element. Describe the order topology of N. 3.4.6 Exercise. Let X = {1, 2} × N have the lexicographic order. Then X becomes an ordered set with minimal element. Describe the order topology of X. Is it discrete? 3.4.7 Proposition. Let X be a topological space and B a family of open sets in X. Then B is a basis for the topology of X if and only if, for every open set U ⊂ X, and for every x ∈ U , there is an element B ∈ B such that x ∈ B ⊂ U . t u 3.4.8 Exercise. Show that in Rn the family {B1/n (q)  n ∈ N, q ∈ Qn } is a basis for the usual topology. This last exercise shows that the topology of Rn admits a countable basis. This property is not shared by all topological spaces. An example would be any discrete uncountable space, since necessarily all its onepoint subsets must belong to any basis. The property of having a countable basis is important as we shall see. 3.4.9 Definition. A topological space X is said to be secondcountable if it satisfies the second countability axiom, namely the axiom (2C) The topology of X admits a countable basis. 3.4.10 Proposition. If a topological space is secondcountable, then it is also firstcountable.
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3 Comparison and generation of topologies
Proof: If X satisfies the second countability axiom, then it has a countable basis B. Take x ∈ X and Bx = {B ∈ B  x ∈ B}. Then Bx is a countable neighborhood basis at x. Hence X satisfies the first countability axiom. t u By Exercise 3.4.8, we have the following result. 3.4.11 Theorem. Rn is secondcountable.
t u
3.4.12 Definition. Let X be a topological space and take a subset A ⊂ X. We say that A is dense in X if the closure A = X. A space X is called separable if it contains a countable dense subset. 3.4.13 Exercise. Let Rl denote R with the socalled lower halfopen interval topology or lower limit topology that has as basis the intervals [a, b), a < b. Rl is also known as the Sorgenfrey line and is also denoted by E. Similarly, Ru denotes R with the upper halfopen interval topology or upper limit topology, which has as basis the halfopen intervals (a, b], a < b. (a) For a subset A ⊆ Rl show that a point x lies in the closure of A if and only if there is a sequence (xn ) in A such that xn ≥ x and xn − x → 0. What is the corresponding statement for Ru ? (b) Show that a function f : Rl −→ R (with the usual topology on R) is continuous if and only if it is continuous from the right at each point x, that is, lim f (x + ε) = f (x) where the limit is as ε → 0 with ε > 0. What is the corresponding statement for Ru ? 3.4.14 Exercise. Consider a function f : R −→ R. (a) Assume that f is continuous from the right, namely, limx→a+ f (x) = f (a) for each a ∈ R, where x → a+ means that the limit is taken if x − a > 0. Prove that f is continuous when considered as a function f : Rl −→ R. (b) What can be said about the continuity of f when considered as f : R −→ Rl and f : Rl −→ Rl ? 3.4.15 Exercise. Let Y be an ordered set furnished with the order topology and assume that f, g : X −→ Y are continuous. (a) Show that the set {x  f (x) ≤ g(x)} is closed in X.
49
3.4 Basis of a topology
(b) Let h : X −→ Y be the function given by h(x) = min{f (x), g(x)} . Show that h is continuous. 3.4.16 Example. More generally than 3.4.3(c), if we have a finite family of ordered sets A1 , . . . , An , with order relations 0 ⇔ x = 0,
1 , n ∈ N} , n
is connected (see Figure 6.4); however, it is not locally connected, since for any point a = (0, y) ∈ P, y > 0, any small enough neighborhood is disconnected. Namely, a neighborhood of each of these points in P looks as in Figure 6.5
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6 Connectedness
Figure 6.4 The infinite comb
Figure 6.5 A neighborhood in the infinite comb 6.2.3 Exercise. (a) Prove that if X is locally connected, then for every x ∈ X, its connected component Cx is open. (b) Let X be the topologist’s sine curve given by taking A = {(x, y) ∈ R2  0 < x ≤ 1, y = sin( πx )}, B = {(0, y) ∈ R2  −1 ≤ y ≤ 1}, and X = A ∪ B. (See Figure 6.6.) Prove that X is connected. Is X locally connected?
6.2.4 Remark. By 6.1.33 and 6.2.3(a), we have that if X is locally connected, then Cx is open and closed. On the other hand, Cx ∩ Cy 6= ∅ ⇔ Cx = Cy , because otherwise Cx ∪ Cy would be a connected set that is strictly larger than Cx and Cy . Thus we can define an equivalence relation in X such that x ∼ y ⇔ Cx = Cy and call Λ the set of equivalence classes. For each λ ∈ Λ, take Cλ = Cx for some
121
6.3 Pathconnected spaces
(0, 0)
(1, 0)
Figure 6.6 The topologist’s sine curve x ∈ λ. The subspaces Cλ ⊂ X are called connected components of X. Hence, the connected components of a space X are the maximal connected sets in X. By Remark 6.2.4 and by Theorem 4.4.10, we have the following. 6.2.5 Theorem. If X is a locally connected space, then a Cλ , X= λ∈Λ
where {Cλ  λ ∈ Λ} is the family of connected components de X.
t u
6.2.6 Exercise. If X is locally connected, then the equivalence relation defined in 6.2.4 determines a quotient map X −→ Λ. Prove that Λ with the quotient topology is discrete. Show with an example that if X is not locally connected, then the previous statement is false. (Hint: Take X = Q.) 6.2.7 Exercise. Show that the Sorgenfrey line E (see 3.4.13) is not locally connected. 6.2.8 Exercise. Show that the continuous image of a locally connected space need not be not locally connected. 6.2.9 Exercise. Show that the nonempty product space if and only if the following hold:
Q
λ∈Λ Xλ
(a) Each factor space Xλ is locally connected. (b) All but finitely many factor spaces Xλ are connected.
is a locally connected
122
6.3
6 Connectedness
Pathconnected spaces
There is a stronger concept than that of connectedness, namely the socalled path connectedness† . 6.3.1 Definition. A path in a topological space X is a map σ : I −→ X. The point x0 = σ(0) is called the start of the path and the point x1 = σ(1), is called the end of the path. We shall denote this fact by σ : x0 ' x1 and we shall also say that σ joins x0 with x1 or that x0 and x1 are connected by the path σ. The following are paths which play an important role. The constant path κx : x ∼ x given by κx (t) = x for all t ∈ I; given a path σ : x0 ' x1 , the inverse path σ : x1 ' x0 given by σ(t) = σ(1 − t); and given two paths σ : x0 ' x1 and τ : x1 ' x2 , the path product στ : x0 ' x2 , defined by ½ σ(2t) if t ≤ 12 , στ (t) = τ (2t − 1) if 21 ≤ t, is a path from x to z.) x σ y
τ
z
σ∗τ
Figure 6.7 The path product
Using the constant, the inverse, and the product paths, one easily proves the next. 6.3.2 Proposition. The relation x0 ∼ x1 ⇔ σ : x0 ' x1 , for some path σ, is an equivalence relation. t u 6.3.3 Definition. The equivalence classes under the relation ' are called the pathcomponents of the space X and the set of equivalence classes, that is the set of pathcomponents, is usually denoted by π0 (X). We say that a topological space X 6= ∅ is path connected if it has only one pathcomponent, that is, if given any two points x, y ∈ X, then there exists a path that joins them. † I like to call the “connectedness” concept, “topological connectedness”, and the “path connectedness” concept, “homotopical connectedness”.
6.3 Pathconnected spaces
123
6.3.4 Examples. The following are pathconnected spaces: (a) The interval I, and with it, any other interval in R, including R itself. (b) Any indiscrete space. (c) The Sierpinski space (2.6.10). 6.3.5 Exercise. Check that in fact the previous are examples of pathconnected spaces. Clearly, the property of a space of being path connected is a topological invariant. 6.3.6 Proposition. Every pathconnected space X is connected. Proof: If X is not connected and x, y ∈ X are in different connected components, say Cx and Cy , and if σ : I −→ X is a path that joins x and y, then the inverse images σ −1 Cx and σ −1 Cy build a disconnection of I. This contradicts 6.1.11(a). t u 6.3.7 Example. The converse of 6.3.6 does not hold. For instance, the topologist’s sine curve defined in 6.2.3(b), and given by ½³ ¾ ³ π ´´ ¯¯ ¯ X= x, sin ¯ x ∈ (0, 1] x is connected, because it is the closure of a connected set. But it is not path connected, since there is no path that joins (0, 0) and (1, 0) in X. See Figure 6.6. Similarly to 6.1.12, we have the following result. 6.3.8 Theorem. Let f : X −→ Y be continuous and let X be path connected. Then f (X) is path connected. Proof: Given f (x), f (y) ∈ f (X), take σ : x ' y; then f ◦ σ : f (x) ' f (y).
t u
¡ ¡ ¢¢ 6.3.9 Remark. The space { x, sin πx  x ∈ (0, 1]} is path connected, since it is ¡ ¡ ¢¢ the image of the interval (0, 1] under the continuous map given by x 7→ x, sin πx . Thus Example 6.3.7, in contrast with 6.1.22, also shows that the closure of a pathconnected space need not be path connected.
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6 Connectedness
Similarly to 6.1.15 and 6.1.16, we have the following two results. 6.3.10 Theorem. Let X be a topological space and for each λ ∈ Λ, take a path S connected subset Xλ ⊂ X such that X = Xλ . If there exists λ0 ∈ Λ such that Xλ0 ∩ Xλ 6= ∅, then X is path connected. Proof: Take two points x, y ∈ X and assume that x ∈ Xλ and y ∈ Xµ . Since Xλ ∩ Xλ0 6= ∅ and Xµ ∩ Xλ0 6= ∅, we may take points x0 ∈ Xλ ∩ Xλ0 and y 0 ∈ Xµ ∩ Xλ0 . Since Xλ and Xµ are path connected, there are paths σ : x ' x0 and τ : y 0 ' y. Furthermore, since Xλ0 is path connected, there exists a path γ : x0 ' y 0 . The path product (σγ)τ : x ' z is defined and joins x and z. Thus X is path connected. t u 6.3.11 Corollary. Let X be a topological space and let Xλ ⊂ X be a family of S path connected sets, where λ ∈ Λ, such that X = Xλ . If ∩Xλ 6= ∅, then X is path connected. t u 6.3.12 Exercise. Let Xn be a sequence of path connected subspaces of X such S that for all n, Xn ∩ Xn+1 6= ∅. Show that n Xn is path connected. The pathconnectedness property is inherited by products. We have the next. 6.3.13 Proposition. If Xλ , λ ∈ Λ, is a nonempty family of nonempty spaces, Q then their product λ∈Λ Xλ is path connected if and only if each of its factors Xλ is path connected. Q Proof: Assume that for every λ, Xλ is path connected, and let x, y ∈ λ∈Λ Xλ be any two points. If xλ , yλ ∈ Xλ are their components in the factor Xλ , since this factor space is path connected, there is a path σλ : I −→ Xλ that joins them. By the universal property of the product, there is a (unique) path σ : I −→ Q λ∈Λ Xλ such that composed with the projection on the factor Xλ one obtains σλ . Therefore, σ : x ' y. Q Conversely, if λ∈Λ Xλ is path connected, since for each λ ∈ Λ, Xλ is the image under the corresponding projection, then by Theorem 6.3.8, Xλ is path connected. t u Consequently, any Euclidean space Rn is path connected. More generally, one has the following.
6.4 Locally pathconnected spaces
125
6.3.14 Examples. 1. Any Euclidean ball, that is, any ball around some point of Rn is path connected. If the ball is open, then this holds simply because this ball is homeomorphic to the space Rn . If it is closed, it is quite obvious that any boundary point can be joined by a path to any interior point (just take the linear path, namely the straight line segment between both points). 2. More generally, any convex set C ⊂ Rn (see 2.6.16) is path connected. Namely, if x, y ∈ C, then the line segment [x, y] ⊂ X. Hence the linear path λ : I −→ C given by λ(t) = (1 − t)x + ty is a path that joins x and y in C.
6.4
Locally pathconnected spaces
As in the case of connectedness, pathconnectedness has a local version. 6.4.1 Definition. We say that a topological space X is locally path connected if each point x ∈ X has a local basis consisting of pathconnected neighborhoods. 6.4.2 Example. Any Euclidean space Rn is locally path connected, since each point has a local basis consisting of balls, that are always path connected (see 6.3.14). 6.4.3 Proposition. Let X be a topological space. The following are equivalent: (a) X is locally path connected. (b) For every open set A ⊂ X, the pathcomponents of A are open. (c) X has a basis consisting of open pathconnected sets. (d) If A ⊂ X is open and q : A −→ A0 is a quotient map that identifies each pathcomponent of A in a point, then A0 is discrete. Proof: (a) =⇒ (b) If A ⊂ X is open and x ∈ A, then by (a) there exists V ∈ Nx such that V ⊂ A and V is pathconnected. Therefore, V ⊂ cx (A), where cx (A) denotes the pathcomponent of A where x lies. Hence, this pathcomponent is open. (b) =⇒ (c) Let B be the family of pathcomponents of open sets of X. By (b) they build a basis for the topology of X.
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6 Connectedness
(c) =⇒ (a) Let B be a basis for X, whose elements are path connected. If x ∈ X, then Bx = {V ∈ B  x ∈ V } is a local basis of pathconnected neighborhoods at the point x. Therefore, X is locally pathconnected. (b) =⇒ (d) Let A be open in X and let q : A −→ A0 be the quotient map that sends each pathcomponent of A to a point. Hence, for x0 ∈ A0 , q −1 (x0 ) is a pathcomponent of A, and thus, by (b) it is open. Therefore, since q is an identification, the singular set {x0 } is open, and thus A0 is discrete. (d) =⇒ (b) Let A be open in X and let q : A −→ A0 be the quotient map that identifies each pathcomponent of A in a point. By (d), A0 is discrete, and so each singular set {x0 } ⊂ A0 is open. Therefore, each pathcomponent of A, cx (A) = q −1 (q(x)), is open. t u 6.4.4 Corollary. If X is locally path connected, then for each x ∈ X the pathcomponent of X, cx , where x lies, is open and closed in X. Proof: By 6.4.3(b), cx is open. Since X − cx = that cx is also closed.
S
y6∈cx cy ,
this set is also open, so t u
Now we can establish a converse result of 6.3.6. Example 6.3.7 shows that not every connected space is path connected. However, the space X of that example is not locally path connected. In fact, it is not locally connected. 6.4.5 Corollary. If X is a connected and locally pathconnected space, then it is a pathconnected space. Proof: If x ∈ X, then the pathcomponent of X in x, cx , is nonempty, open, and closed. Therefore, since X is connected, cx = X, and thus X is path connected. t u To finish this section, we shall give some examples. 6.4.6 Examples. (a) The space X of 6.2.3 is connected, but it is not locally path connected, and it is not either path connected. However, if C = {(0, y)  y ∈ [− 32 , −1]} ∪ {(x, − 23 ) ∈ R2  x ∈ [0, 1]} ∪ {(1, y)  y ∈ [− 32 , 0]}, then the space Y = X ∪ C is path connected, but it is not locally path connected. This space Y is known as the Polish circle (see Figure 6.8). (b) The infinite comb, defined in 6.2.2(c) as the subspace of R2 given by P = {(x, y) ∈ R2  y = 0, or x = 0, n1 ; n = 1, 2, . . . and 0 ≤ y ≤ 1}, is path connected, but not locally (path) connected.
6.4 Locally pathconnected spaces
127
Figure 6.8 The Polish circle 6.4.7 Exercise. Prove that if X is a locally pathconnected space, then for every x ∈ X the connected component Cx coincides with the pathcomponent cx . 6.4.8 Exercise. (a) Determine the connected components and the path components of the space Rω as defined in Example 6.1.26. (b) Show that x, y ∈ Rω lie in the same component of Rω if and only if the sequence x − y is eventually zero. (Hint: Prove that if x − y is not eventually zero, then there is a homeomorphism ϕ : Rω −→ Rω such that ϕ(x) is bounded and ϕ(y) is unbounded.) 6.4.9 Exercise. Consider in R2 the set IQ of all rational points in the unit interval I and let R be the set {1} × IQ . Define X as the union of all line segments that join the origin with each of the points of R. (a) Show that X is path connected. (b) Show that the only point at which X is locally path connected is the origin. 6.4.10 Exercise. Consider in R2 the set IQ of all rational points in the unit interval I and let X be the union of all line segments {q} × I, q ∈ IQ with I × {0}. (a) Show that X is path connected.
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6 Connectedness
(b) At which points is X locally path connected? 6.4.11 Exercise. Consider in R2 the set K = { n1  N} and let S be the set {1} × K. Define X as the union of all line segments that join the origin with each of the points of S. (a) Show that X is path connected. (b) At which points is X locally path connected? 6.4.12 Exercise. (a) Show that the closure of the topologist’s sine curve X = {0} × [−1, 1] ∪ {(t, sin( πt ))  t ∈ I} (see 6.2.3(b)) is not path connected. (Hint: If σ : I −→ X is a path with start at the origin and end in {(t, sin( πt ))  t ∈ I}, then the set of those t for which σ(t) ∈ {0} × [−1, 1] is closed, thus it has a largest element b. Hence the restriction λ : [b, 1] −→ X maps b into {0} × [−1, 1] and (b, 1] into {(t, sin( πt ))  t ∈ I}. For each t ∈ [b, 1], put λ(t) = (x(t), y(t)). π Thus x(b) = 0 and x(t) > 0 and y(t) = sin( x(t) ) for t > b. Show that there is a sequence of points tk → b such that y(tk ) = (−1)k , thus not convergent. This contradicts the continuity of σ.) (b) Conclude that the polish circle (see 6.4.6) is not locally path connected.
Chapter 7
Filters
Filters are a concept defined on sets. They constitute a powerful tool to prove fundamental results in topology. Among other things, their good behavior with respect to cartesian products allows to use them to prove theorems which state that products have certain properties if and only if their factors have them. One important example of this fact is the Tychonoff theorem on compactness, which making use of filters we shall prove in the next chapter. Filters codify and extend in some way the behavior of sequences. In the topological setup one has a concept of convergence of filters. On the other hand, the set of neighborhoods of a point are an example of a filter. In this sense, filters are a common generalization of both sequences and neighborhoods, which is very useful to better understand both concepts. Filters allow to study convergence in arbitrary topological spaces. These properties reduce to properties of sequenceconvergence in firstcountable spaces, i.e. spaces for which each point has a countable neighborhood basis. Such properties of filters can be inferred from the analysis of sequence convergence.
7.1
Filters
In this section we shall introduce the concepts of a filter basis and of a filter. We shall study the relationships between different filters, as well as of maximal filters which will be called ultrafilters. Making use of filters we shall generalize the concept of sequence convergence and we shall reformulate classical concepts about convergence in metric spaces. There is a more detailed treatment of filters in other texts such as [15], [20], or [7]. We start recalling some concepts already defined in the case of metric spaces in Section 1.4. 7.1.1 Definition. A sequence in a topological space (or in a set) X is a function N −→ X, n 7→ xn . This function is denoted by (xn ) or simply by xn . We say that 129
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the sequence converges to x, in symbols xn → x, if for every neighborhood V of x there is a number N ∈ N such that xn ∈ V if n ≥ N . In this case, we say that x is a limit of the sequence. For each N ∈ N we call tail of (xn ) the restriction (xn )n>N . Hence a sequence converges to a point if for each neighborhood of this point there is a tail of the sequence which is contained in this neighborhood. We start observing the following. 7.1.2 Note. Assume that a space X is firstcountable (see 2.4.3), and let {Un } be a countable neighborhhood basis at x ∈ X. We can always assume that for all n, Un ⊇ Un+1 for all n. Such a neighborhood basis is said to be nested. If the given basis is not already nested, we may replace it by the neighborhood basis {Un0 }, where Un0 = U1 ∩ · · · ∩ Un . The next is an easy exercise. 7.1.3 Exercise. Assume that {Un } is a nested countable neighborhood basis at x ∈ X, and for each n take any point xn ∈ Un Show that xn → x. The following result characterizes the topology of a firstcountable space, as well as the continuity of a map, whose domain is firstcountable, using sequence convergence. 7.1.4 Theorem. Let X be a firstcountable topological space. Then the following hold: (a) A ⊂ X is closed if and only if for every sequence (xn ) in A such that xn → x in X, one has that x ∈ A. (b) f : X −→ Y is continuous at x ∈ X if and only if for every sequence (xn ) in X such that xn → x, the image sequence f (xn ) → f (x). Proof: (a) Assume first that A ⊂ X is closed. Let (xn ) be a sequence of elements of A such that xn → x in X. If x ∈ / A, then X − A is an open neighborhood of x and thus there is a natural number N such that xn ∈ X − A for all n ≥ N . This is a contradiction, since xn ∈ A for all n. Conversely, take an arbitrary point x ∈ A. Since X is firstcountable, there is a nested neighborhood basis of x, say {Un }. Since Un ∩ A 6= ∅, we may take a point
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xn ∈ A ∩ Un , and since the neighborhood base is nested, it follows by Exercise 7.1.3 that xn → x. Therefore, by assumption, x ∈ A. Consequently A ⊂ A and therefore A is closed. (b) Assume first that f is continuous at x and suppose xn → x. Take a neighborhood V of f (x). Since f is continuous at x, there is a neighborhood U of x such that f (U ) ⊆ V . Furthermore, since xn → x, there is an N such that if n ≥ N , then xn ∈ U . Therefore, if n ≥ N , then f (xn ) ∈ f (U ) ⊆ V and thus, f (xn ) → f (x). Conversely we shall prove that under the assumption, for all A ⊆ X, f (A) ⊂ f (A). Thus by Theorem 2.5.8 (d), f shall be continuous. So take an arbitrary point x ∈ A. Since X is firstcountable, there is a nested neighborhood basis of x, say {Un }. On the other hand, Un ∩ A 6= ∅, so we may take a point xn ∈ A ∩ Un . Since the neighborhood base is nested, once again it follows by Exercise 7.1.3 that xn → x. Hence, by assumption f (xn ) → f (x), and since f (xn ) ∈ f (A) for all n, f (x) ∈ f (A). t u 7.1.5 Note. Observe that in the first part of the proofs of (a) and (b) in the previous theorem, the assumption that X is first countable is not necessary. But for the second parts the assumption is needed as the following exercise shows. 7.1.6 Exercise. Consider an uncountable set X with the cocountable topology (2.1.2(f)) on it, namely, A ⊂ X is open if and only if either A = ∅ or X − A is at most countable. Show that a sequence in X is convergent if and only if the sequence is finally constant, i.e. it has a constant tail. Hence any function f : X −→ Y will map convergent sequences in convergent sequences. However not any function f must be continuous. Give examples of Y and f such that f is not continuous. Show explicitly that X is not firstcountable. In general, as we saw above, without the firstcountability assumption, Theorem 7.1.4 is not true. Therefore, we must generalize the concept of sequence convergence. To do that, consider the following analogy between sequences and neighborhoods of a point: The intersection of two tails of a sequence is again a tail. Analogously, the intersection of two neighborhoods of a point is again a neighborhood of that point. This analogy can be extended to basic neighborhoods, observing that the intersection of two of them contains another. These observations suggest the next definition. 7.1.7 Definition. A nonempty system B of nonempty subsets of a set X is called filter basis in X if the following holds:
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(FB) B1 , B2 ∈ B =⇒ ∃B ∈ B such that B ⊂ B1 ∩ B2 . 7.1.8 Examples. (a) If A ⊂ X is nonempty, then the set {A} is a filter basis. (b) Let (xn ) be a sequence in the set X. Then the system B of tails of (xn ) is a filter basis in X. (c) Let X be topological space and take x ∈ X. A system B of basic neighborhoods in x is a filter basis in X. (d) Let Xλ , λ ∈ Λ, be a nonempty family of nonempty sets. If Bλ is a filter basis in Xλ , then the products Y Xλ , Bl1 × · · · × Blk × λ6=l1 ...lk
such that Blj ∈ Blj , j = 1, . . . , k, constitute a filter basis in
Q
Xλ .
7.1.9 Definition. A nonempty system F of nonempty subsets of a set X is filter in X if (F1) F ∈ F, F 0 ⊃ F =⇒ F 0 ∈ F. (F2) F1 , F2 ∈ F =⇒ F1 ∩ F2 ∈ F. Notice that necessarily X ∈ F for every filter F. 7.1.10 Examples. (a) Given a filter basis B in a set X, the family of all supersets of the sets in B constitutes a filter F, called the filter generated by B. Namely F = {F ⊆ X  ∃B ∈ B such that B ⊆ F } . We shall say that B is a basis of F. (b) Let X 6= ∅, then {X} is a filter in X called the trivial filter. It is the filter generated by the filter basis {X}. (c) All supersets of nonempty set A ⊂ X build a filter denoted by FA , namely FA = {F ⊆ X  A ⊆ F }. It is the filter generated by the filter basis {A}.
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(d) Given a sequence (xn ) in a set X, the supersets of the tails of the sequence form a filter called elementary filter generated by the sequence. It is denoted by F(xn ) . Namely F(xn ) = {F ⊆ X  ∃N ∈ N such that xn ∈ F for all n > N } . It is the filter generated by the filter basis of tails of the sequence. (e) Let X be a topological space and take x ∈ X. The set Nx of all neighborhoods of x is a filter in X called neighborhood filter of x ∈ X. This is always a filter generated by a neighborhood basis, seen as a filter basis. (f) Let X be an infinite set. The system of cofinite sets of X, namely sets whose complement is finite, is a filter in X called cofinite filter. (g) Every filter is a filter basis. 7.1.11 Definition. We shall say that two filter bases are equivalent if both generate the same filter according to Example 7.1.10(a). 7.1.12 Example. Consider B = {(−1/n, 1/n) ⊂ R  n ∈ N} and take B 0 = {(−π/2n , π/2n ) ⊂ R  n ∈ N}. Both sets are filter bases and they are equivalent since they generate precisely the neighborhood filter of 0 in R. As we saw in Example 7.1.10(d), every sequence determines an elementary filter. Moreover different sequences may determine the same filter, for instance sequences only differ in a finite number of their terms. If we observe the definition of convergence of a sequence in a topological space, we see this convergence is based in a comparison of the tails of the sequence and the neighborhoods of the point to which the sequence converges. More precisely, the sequence (xn ) converges to a point x if for each neighborhood of x there is a tail of the sequence contained in the neighborhood. In other words, the sequence (xn ) converges to x if each neighborhood of x is an element of the elementary filter generated by the sequence. Or in symbols, xn → x if Nx ⊂ F(xn ) . This suggests the next concept. 7.1.13 Definition. Let F and G be two filters in a set X. If F ⊂ G, then we say that F is coarser than G and that G is finer than F. 7.1.14 Examples. (a) The trivial filter {X} is the coarsest filter in X.
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(b) There is no finest filter of all. Namely, if A and X − A are nonempty, then each of them determines a superset filter. Namely the filters FA y FX−A . There can be no filter finer than both, then otherwise this filter would have A and X − A as elements, and hence also their intersection A ∩ (X − A) = ∅, which is impossible. 7.1.15 Theorem. Let B1 and B2 be filter bases in the set X. The following are equivalent: (a) The filter generated by B1 is finer than the filter generated by B2 . (b) Each set in B2 contains one set in B1 .
t u
7.1.16 Corollary. Two filter bases B1 and B2 in a set X are equivalent if and only if each set in B1 contains one set in B2 and viceversa. t u Notice that Example 7.1.12 illustrates quite well the statement of the previous corollary. From the considerations previous to Definition 7.1.13, the following concept is quite natural. 7.1.17 Definition. Let X be a topological space and let F be a filter in X. We say that the filter F converges to a point x ∈ X, in symbols F → x, if the filter F is finer than the neighborhood filter of x. I.e. F → x if Nx ⊂ F. If B is a filter basis in X, then we say that B converges to x ∈ X, in symbols B → x, if the filter generated by B converges to x. We write lim F for the set of all points x ∈ X such that F converges to x, i.e. in symbols lim F = {x ∈ X  F → x}. We call this the limit ofF. If lim F has only one point x, then we write lim F = x. If lim F = ∅ then we say that F does not converge or simply that it diverges. 7.1.18 Proposition. Let a filter G be finer than a filter F. If F converges to x, then G converges to x too. t u 7.1.19 Examples. (a) Let X be a topological space and take a sequence (xn ) in X. Then (xn ) converges to x if and only if the elementary filter generated by (xn ) converges to x. (b) Let X be an indiscrete space. Then for every filter F and for every point x in X, F → x, i.e. lim F = X.
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7.1.20 Proposition. Let {Fλ }λ∈Λ be a family of filters in a set X. Then the T intersection F = Fλ is a filter. We call this filter simply the intersection of the filters Fλ , or infimum of {Fλ }. We denote it by inf Fλ . If G is a filter in X which is coarser than Fλ for all λ ∈ Λ, then G ⊂ Fλ for all λ ∈ Λ. Clearly G is also coarser than inf Fλ , namely, inf Fλ is the finest of all filters that are coarser than all Fλ . 7.1.21 Theorem. Let X be a topological space. The infimum of all filters which converge to x ∈ X is the neighborhood filter Nx . t u Let now {Bλ }λ∈Λ be a family of filter bases in a set X and let Fλ be the filter generated by Bλ . If there is a filter which is finer than all filters Fλ , then it must contain all sets of all the Fλ , or of all Bλ . It must also contain all finite intersections of sets in the filter bases Bλ . This suggests to define B={
n \
Bi  Bi ∈ Bλi , λi ∈ Λ} ,
i=1
and we clearly have the next.
7.1.22 Proposition. If no set in B is empty, then B is a filter basis.
t u
Under the assumption of 7.1.22, B is a basis of the coarsest filter which is finer than all filters Fλ . Thus we have the following result. 7.1.23 Theorem. Let {Bλ }λ∈Λ be a family of filter bases in a set X. Then there is a finer filter than all filters generated by the filter bases Bλ if and only if the T finite intersections ni=1 Bi , where Bi ∈ Bλi , λi ∈ Λ (different), are nonempty. If this is the case, then these intersections constitute a basis of the coarsest filter which is finer than all filters generated by the filter bases Bλ . t u 7.1.24 Definition. Let {Bλ }λ∈Λ be a family of filter bases in a set X. Then the T filter F generated by the filter basis B = { ni=1 Bi  Bi ∈ Bλi , λi ∈ Λ} is called supremum of the filters Fλ generated by the Bλ . In symbols, F = sup Fλ . 7.1.25 Theorem. Let X be a set and take A ⊂ X. Let F be a filter in X. Then A ∈ F 0 for some filter F 0 finer than F if and only if X − A 6∈ F.
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Proof: If X − A ∈ F, then X − A ∈ F 0 for all F 0 finer than F. Hence A 6∈ F 0 for all F 0 finer than F. Conversely, if X − A 6∈ F and F ∈ F, then F 6⊂ X − A. Hence F ∩ A 6= ∅. By the previous theorem, the supremum of F and the filter FA of all supersets of A must exist, i.e. there would exist a filter F 0 , which is finer than F and has A as an element. t u 7.1.26 Exercise. Let B1 and B2 be filter bases. Show that B = {B1 ∪ B2  B1 ∈ B1 and B2 ∈ B2 } is a filter basis. What relationship keeps the filter generated by B with the filters generated by B1 y B2 , respectively? We now consider the following question. Which property must X have in order that lim F ⊆ {x}, x ∈ X? That is, under what condition on X does a filter have at most one limit point. This requires that there is no filter finer than the neighborhood filters of any two points of X. We shall study this condition in detail in what follows. Let X be a topological space, take points x 6= y in X. There is a finer filter than the neighborhood filter of x, Nx , and the neighborhood filter of y, Ny , if and only if every neighborhood of x meets all neighborhoods of y. In other words, there is a filter F such that F → x and F → y if and only if there is a filter F such that Nx ⊂ F and Ny ⊂ F. This is equivalent, as we said above, to the fact that each neighborhood of x intersects each neighborhood of y. Thus we have proved the following. 7.1.27 Proposition. Any filter in X converges to at most one point if and only if the following condition holds: (H) For each pair of points x 6= y in X, there are neighborhoods U ∈ Nx and V ∈ Ny such that U ∩ V = ∅. t u Condition (H) is known as Hausdorff separability axiom. 7.1.28 Proposition. Axiom (H) is equivalent to the following: (H0 ) Each point of X is the intersection of all its closed neighborhoods. Proof: (H)=⇒(H0 ) Let x ∈ X be an arbitrary point of X. For each point y 6= x, there are open disjoint neighborhoods U and V of x and y, respectively. Hence
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W = X − V ⊃ U , so that W is a closed neighborhood of x that does not contain T y as an element. Therefore {x} = {W  W is a closed neighborhood of x}. (H0 )=⇒(H) Take x, y ∈ X, x 6= y. There is a closed neighborhood V of y such that x 6∈ V . Hence X − V is an open set that contains x as an element. Thus U = X − V is a neighborhood of x which does not intersect V . t u 7.1.29 Definition. The set ∆ = {(x, x) ∈ X × X} is called the diagonal of the space X in X × X. Clearly (x, y) 6∈ ∆ if and only if x 6= y. If X satisfies (H) and (x, y) 6∈ ∆, then there are disjoint neighborhoods U and V of x and y, respectively. The product U × V is a neighborhood of (x, y) in X × X such that (U × V ) ∩ ∆ = ∅, since if a diagonal point (z, z) ∈ U × V , then z ∈ U ∩ V . Hence ∆ is closed in X × X. Conversely, if ∆ is closed in X × X, then the difference set X × X − ∆ is open and if x 6= y, i.e. if (x, y) ∈ X × X − ∆, then there is a neighborhood W of (x, y) in X × X, which does not meet ∆. Hence there are neighborhoods U and V of x y y in X, respectively, such that U × V ⊂ W . This way (U × V ) ∩ ∆ = ∅ or equivalently U ∩ V = ∅. Thus we have shown the following result. 7.1.30 Proposition. Axiom (H) is equivalent to the following: (H00 ) The diagonal ∆ ⊂ X × X is closed.
t u
The two previous propositions can be summarized in the next result. 7.1.31 Theorem. Let X be a topological space. The following are equivalent: (H) Any two different points of X have disjoint neighborhoods. (H0 ) Any point of X is the intersection of all its closed neighborhoods. (H00 ) The diagonal in X × X is closed. (H000 ) No filter in X converges to more than one point.
t u
7.1.32 Definition. A topological space X that satisfies one (and thus all) of the axioms (H)–(H000 ), is called Hausdorff space or T2 space. 7.1.33 Examples.
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x U
y V
X
Figure 7.1 Disjoint neighborhoods of different points (a) Every metric space X is Hausdorff. Namely, given x 6= y in X and ε = d(x, y)/2, then Bε (x) ∩ Bε (y) = ∅. (b) Not every pseudometric space is Hausdorff. For instance the indiscrete space with more than one element is not Hausdorff.. The property of being a Hausdorff space is hereditary, i.e. is a property that is inherited by all subspaces. 7.1.34 Proposition. Every subspace of a Hausdorff space is a Hausdorff space, i.e. the Hausdorff separability property is hereditary. t u Since in a Hausdorff space each point is the intersections of all its closed neighborhoods, we have the following. 7.1.35 Proposition. In a Hausdorff space every onepoint set is closed.
t u
The converse of this proposition is false, as the next example shows. 7.1.36 Example. Let X be an infinite set with the cofinite topology (see 2.1.2(e)). It is obvious that each onepoint set is closed. However, the space is not Hausdorff. 7.1.37 Proposition. A topological space X is such that each of its onepoint sets is closed if and only if it satisfies
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(T1 ) For every pair of points x 6= y in X there are neighborhoods U and V of x and y in X, respectively, such that y 6∈ U and x 6∈ V . t u 7.1.38 Definition. Axiom (T1 ) is known as the first separability axiom. A topological space X which satisfies the axiom (T1 ) is called a T1 space. (some authors call these Fr´echet spaces, but this might lead to confusion and we avoid this name). 7.1.39 Exercise. Let X be an infinite set. Show that the coarsest topology on X such that it becomes a T1 space is the cofinite topology. 7.1.40 Exercise. Show that Rn with the Zariski topology (2.2.23) is T1 space, which is not Hausdorff. What relation has this example with 7.1.36? 7.1.41 Note. Axiom (H) is sometimes denoted by (T2 ) and is also is known as the second separability axiom. A topological space X which satisfies axiom (T2 ), i.e. a Hausdorff space, is also called a T2 space. Axiom (T1 ) is weaker than (T2 ), namely (T2 ) =⇒ (T1 ). There is a weaker axiom than (T1 ), namely (T0 ) For every pair of points x 6= y in X there is either a neighborhood U of x in X, or a neighborhood V of y in X, such that y 6∈ U or x 6∈ V . Axiom (T0 ) is known as the zeroth separability axiom. We have (T2 ) =⇒ (T1 ) =⇒ (T0 ). 7.1.42 Exercise. Show in an example that not all T0 space is a T1 space. (Hint: Consider the Sierpinski space.) Exercises 7.1.42 and 7.1.36 or 7.1.40 show that no two of the three separability axioms are equivalent. 7.1.43 Exercise. Let X be a metric space with metric d and let (xn ) be a sequence in X. Show that xn → x if and only if d(xn , x) → 0. To finish this section we shall see how a filter in a set generates a topology on it.
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7.1.44 Example. Let X be a set and F an arbitrary filter in X. Take the set X ∗ = X ∪ {∞} and for x ∈ X ⊂ X ∗ define its neighborhood filter by Nx = Fx , namely the filter of all sets in X ∗ that contain {x}. For ∞ ∈ X ∗ , define the neighborhood filter by N∞ = {F ∪ {∞}  F ∈ F}. It is an exercise to show that these sets make up a neighborhood system on the set X ∗ , and that they determine a topology on X ∗ . Denote the resulting topological space by XF∗ . It is an interesting exercise to consider the special case in which X is an infinite set and F is the cofinite filter described in 7.1.10(f). 7.1.45 Exercise. Show that the subspace X ⊂ XF∗ (with the relative topology) is discrete and is not closed. Conclude that X = XF∗ , namely, X is dense in XF∗ (see Definition 7.4.17).
7.2
Cluster points
Following the parallelism between sequences and filters, the concept of a cluster point of a sequence can be extended to filters. In this section we shall study the concept of cluster point of a filter and what relation it has with comparable filters. We start with the special case of cluster point of a sequence. 7.2.1 Definition. Let X be a topological space and {xn } a sequence of points in X. A point x ∈ X is cluster point de (xn ) if for all U ∈ Nx xn ∈ U for infinitely many values of n. Equivalently x is a cluster point of (xn ) if and only if for all U ∈ Nx , U meets every tail of the sequence (xn ) or, equivalently, x is a limit point of each tail of the sequence. As we already said, the concept of sequence appears to be poor if we consider it in very general topological spaces. The following example due to R. Arens shows the pathological behavior of sequences and justifies the use of of filters as alternative to study convergence. 7.2.2 Example. Let X = (N × N) ∪ {(0, 0)} have the topology such that N × N is discrete and the neighborhoods of (0, 0) are the sets U ⊂ X for which (0, 0) ∈ U and there is an N ∈ N such that for all n ≥ N , U contains all points of {n} × N but finitely many. This topology is Hausdorff. The diagonal sequence, namely, the sequence given by xn = (n, n) has (0, 0) as a cluster point. However it does not contain any subsequence that converges to (0, 0).
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7.2.3 Exercise. Check all statements in the previous example. 7.2.4 Exercise. Let X be a firstcountable space and let (xn ) be a sequence in X with a cluster point x0 . Show that (xn ) has a subsequence that converges to x0 . Starting from the idea that the tails of a sequence form a filter basis, we formulate the following analogy. Let F be a filter in X and let B be a filter basis of F. If x is a limit point of every set in B, then it is a limit point of every set in F. In other words, x ∈ A foe every A ∈ B. If we take B ∈ F, then there is an A ∈ B T such that A ⊂ B. Hence x ∈ B for every B ∈ F. Therefore x ∈ {B ∈ F  B is T closed}. Conversely, it is clear that if x ∈ {B ∈ F  B is closed}, then x ∈ A for every A ∈ B. Thus we have the next definition. 7.2.5 Definition. Let F be a filter in a topological space X. We say that x ∈ X is a cluster point of F if for every U ∈ Nx and for all F ∈ F, U ∩ F 6= ∅. We have the following assertion. 7.2.6 Proposition. A point x is a cluster point of a filter F in a space X if and T only if x ∈ {B ∈ F  B is closed}. t u From the previous proposition, it is clear that the definition of a cluster point of a filter is consistent with that of a cluster point of a sequence (7.2.1). We have the following. 7.2.7 Corollary. A point x in X is a cluster point of a sequence (xn ) if and only if x is a cluster point of the elementary filter F(xn ) defined by the sequence. t u More generally we have the next result. 7.2.8 Theorem. Let X be a topological space and let F be a filter in X. Then x is a cluster point of F if and only if x is a cluster point of F 0 for every filter F 0 coarser than F. t u The relation of the cluster points of a filter with a filter basis that generates the filter is established in the next result, whose proof is immediate. 7.2.9 Theorem. Let X be a topological space and let B be a filter basis in X. Then the following are equivalent:
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(a) x is a cluster point of the filter F generated by B. (b) For each U ∈ Nx and for each B ∈ B, U ∩ B 6= ∅. (c) x ∈ A for all A ∈ B.
t u
We have, in particular, that if A is a nonempty subset of a topological space X, then {A} = B is a filter basis for the filter FA of all supersets of A. Thus analogously to 7.2.7, we get the next. 7.2.10 Proposition. Let X be a topological space and take A ⊂ X. Then x ∈ A, i.e. x is a cluster point of A, if and only if x is a cluster point of the filter FA of all supersets of A. Hence A = {x  x is a cluster point of FA }. t u This last result shows also the consistency of the concept of a cluster point of a filter with that of a set. Theorem 7.2.9 states that the fact that x is a cluster point of the filter F generated by B and the fact that every neighborhood of x meets every set in F are equivalent. Thus the statement is also equivalent to saying that there is a finer filter than both F and Nx . In other words, that there is a finer filter than F which converges to x. Thus we have the following result. 7.2.11 Theorem. Let X be a topological space and let F be a filter in X. Then x is a cluster point F if and only if there is a filter G finer than F such that G converges to x. In particular, if F converges to x, then x is a cluster point of F.u t If a topological space X is not Hausdorff, then a filter might have a limit and several cluster points. 7.2.12 Example. Let X be the Sierpinski space, namely X = {x, y} with the topology A = {X, ∅, {x}}. The filter F = {X} converges to y, however x is a cluster point of F, although F does not converge to x. We have the following result. 7.2.13 Theorem. If X is a Hausdorff space, then for each convergent filter F, the only point in lim F is the only cluster point of F. Proof: Let x = lim F and let y be a cluster point of F. Hence there is a filter G finer than both F and Ny . Thus G → y. But clearly G → x too. Since X is Hausdorff, x = y. t u
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143
7.2.14 Exercise. Let B = {(a, ∞) ⊂ R  a ∈ R}. (a) Show that B is a filter basis. The filter F generated by B is called Fr´echet filter in R. (b) Show that the Fr´echet filter in R with the usual topology does not have cluster points. (c) Let L be the wrapped line (v´ease 2.4.7). Show that the Fr´echet filter in L converges to 0. Similarly to sequences, a filter might have no cluster point or just one cluster point without being convergent. 7.2.15 Exercise. Give the details of the proof of 7.2.7, namely, of the fact that x is a cluster point of a sequence (xn ) if and only if x is a cluster point of the elementary filter F(xn ) associated to {xn }. Analogously prove that x = lim xn if and only if x = lim F(xn ) . Take a firstcountable space X. In terms of filters, this means that for all x ∈ X, the neighborhood filter Nx admits a countable basis. We may consider any filters which admit a countable basis. 7.2.16 Exercise. If F has a countable basis, then then it has a nested basis B = {Bn }n∈N , namely a filter basis such that if m > n, then Bm ⊂ Bn . 7.2.17 Theorem. If F admits a countable basis, then there is an elementary filter which is finer than F. Moreover, F is the intersection of all elementary filters that are finer than F. Proof: Let B = {Bn } be a nested countable basis of F (as in 7.2.16) and take xn ∈ Bn . It is clear that the elementary filter F(xn ) is finer que F. T Conversely, take G = {F(xn )  F(xn ) ⊃ F }. Clearly, G is finer than F. If G= 6 F, then there is a set A ∈ G such that A 6∈ F. Thus A 6⊃ Bn for all n ∈ N, i.e. Bn ∩ (X − A) 6= ∅. Take any xn ∈ Bn ∩ (X − A). Obviously F(xn ) is finer than F However A 6∈ F(xn ) , since {xn } ⊂ X − A. Thus X − A ∈ F(xn ) . This is a contradiction, hence G = F. t u
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The theorem above states, in particular, that if a point x ∈ X has a countable neighborhood basis, then its neighborhood filter Nx is determined by all sequences in X that converge to x and, by definition of filter convergence, all filters that converge to x are also determined by the same sequences. However one should notice and take care that among all filters converging to x, filters may exist which do not admit a countable basis. Namely, if F ⊂ G and G admits a countable basis, this does not mean that F admits countable basis. 7.2.18 Example. The filter F{x} of all supersets of x is finer than its neighborhood filter Nx . A base for F{x} is {x}, which is (at most) countable). However, if X is not firstcountable, the filter Nx might not admit a countable basis. (It is also possible that if a filter does not admit a countable basis, there could be a coarser filter which does admit it.) 7.2.19 Theorem. Let X be a first countable space. Let F be a filter in X which admits a countable basis and let x be a cluster point of F, then there is an elementary filter finer than F which converges to x. Proof: Let B = {Bn } be a basis of F and let U = {Un } be a basis of Nx . We may assume that these bases are nested (7.2.16). If x is a cluster point of F, then Bn ∩ Un 6= ∅ for all n. Take xn ∈ Bn ∩ Un . Then clearly the elementary filter F(xn ) satisfies the statement of the theorem. t u Notice that, in particular, if FA is the filter of all supersets of A 6= ∅ in a firstcountable space X, then x ∈ A if and only if there is a sequence (xn ) in A such that xn → x. Thus we have that for firstcountable spaces, sequenceconvergence allows us to characterize their topology, see 7.1.4. 7.2.20 Exercise. Let X be an uncountable set and take a sequence (xn ) in X, whose points are all different. Show that the filter consisting of all cofinite sets in X (i.e. whose complement is finite) is strictly coarser than the elementary filter F(xn ) and does not allow a countable basis. 7.2.21 Exercise. Let X be a set. (a) Show that the intersection of two elementary filters in X is an elementary filter. (b) Show that the intersection of two filters in X, which admit a countable basis, admits a countable basis too.
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145
(c) Show that if the supremum of a countable family of filters exists, then this supremum admits a countable basis. 7.2.22 Exercise. Let X be a topological space and let B be the filter basis consisting of one nonempty set A ⊂ X. Show that if A = {x0 }, then B → x0 . 7.2.23 Exercise. Let X be a topological space and let F be a filter in X. Show that the set of cluster points of F is closed (possibly empty). 7.2.24 Exercise. Let X be a Hausdorff space and let F be a filter in X such that F → x0 . If G is finer than F and x is a cluster point of G, then show that x = x0 and G → x0 .
7.3
Ultrafilters
In this section we shall study the class of ultrafilters, which are the maximal elements in the class of all filters in a set, with respect to the order relation introduced in Section 7.1. They are filters filters which have particularly interesting properties. We start with a general definition. 7.3.1 Definition. Let M be a set. We say that M is a partially ordered set, also briefly called poset, if it has an order relation or a partial order ≤ which fulfills for any a, b, c ∈ M the axioms (OR1), (OR2), and (OR3) of Definition 3.4.2. 7.3.2 Examples. The following are examples of partially ordered sets: (a) Let X be a set. Then M = P(X), the power of X, with the relation ≤ = ⊆ is a poset. (b) Let X be a set. Then M = {A  A is a topology on X}, with the relation ≤ = “be coarser than” is a poset. (c) Let X be a set. Then M = {F  F is a filter in X} with the relation ≤ = ⊆ is a poset. (d) M = R and Z the usual order relation are posets.
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7.3.3 Definition. Let M be a partially ordered set. An element a ∈ M is called maximal if for no b 6= a, a ≤ b. 7.3.4 Examples. In Examples 7.3.2 the following elements are maximal: (a) X ∈ M . (b) The discrete topology on X. (c) The filters that do admit filters which are strictly finer. (d) There are no maximal elements in R nor in Z. 7.3.5 Definition. Let M be a partially ordered set and take A ⊂ M . An upper bound of A is an element b ∈ M such that a ≤ b for all a ∈ A. If such an upper bound of A exists, then A is said to be bounded from above. 7.3.6 Examples. In Examples 7.3.2 we have the following: (a) y (b) Every subset of M is bounded from above. (c) Not every set of filters is bounded from above. (d) Not every subset is bounded from above, for instance M itself. However, we have the following result. 7.3.7 Theorem. Let M be the set of all filters in a set X with the order relation ⊆. Each totally ordered subset of M has an upper bound. S Proof: Let A be a totally ordered set of filters in X and take F0 = F ∈A F. It is straightforward to verify that F0 is a filter, which is obviously an upper bound of A. t u The following is an important result in Set Theory, whose proof we omit (see [9]). 7.3.8 Theorem. (Zorn’s lemma) Let M be a nonempty partially ordered set. If each totally ordered subset of M has an upper bound, then for every a ∈ M there is a maximal element b ∈ M such that a ≤ b. t u 7.3.9 Definition. A maximal filter in a set X is called ultrafilter.
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As a consequence of 7.3.7 and 7.3.8, we have the following result. 7.3.10 Theorem. Let X be a set. Then for each filter F in X, there is an ultrafilter U which is finer than F. t u 7.3.11 Example. Let X be a set and take x ∈ X. Then the filter Fx = {A ⊂ X  x ∈ A} is an ultrafilter. Indeed, these are the only ultrafilters which can be explicitly described. 7.3.12 Example. Let X a set and take an infinite sequence (xn ). Then the associated elementary filter F(xn ) is not an ultrafilter, since all proper subsequences determine elementary filters, which are strictly finer than F(xn ) . We saw in Theorem 7.2.17 that for every filter F which has a countable basis, there is an elementary filter F 0 finer than F. We also observed in the previous example that the elementary filters are almost never ultrafilters. Consequently, the ultrafilters are either as Fx or do not admit countable bases. We have the following. 7.3.13 Theorem. Let X be a set. Then a filter U in X is an ultrafilter if and only if for each A ⊂ X one has either A ∈ U or X − A ∈ U. Proof: Assume first that U is an ultrafilter. If X − A 6∈ U, then by 7.1.25 we know that there is a filter F which is finer than U such that A ∈ F. But since U is an ultrafilter F = U. Conversely, if U is not an ultrafilter, then there is an ultrafilter F, which is finer than and different from U. Take A ∈ F − U. Clearly A, X − A 6∈ U, since if X − A ∈ U ⊂ F, then we would have X − A, A ∈ F, which is impossible. t u 7.3.14 Corollary. Let X be a set and let U be an ultrafilter in X. If A1 , . . . , An ⊂ S X are such that ni=1 Ai ∈ U, then Ai ∈ U for some i ∈ {1, . . . , n}. Proof: If Ai 6∈ U for all i, then by 7.3.13, X − Ai ∈ U for all i and therefore n \
(X − Ai ) = X −
i=1
which is a contradiction.
n [
Ai ∈ U ,
i=1
t u
If in the previous corollary we take A1 = A and A2 = X − A, then we recover Theorem 7.3.13. Hence 7.3.13 and 7.3.14 are equivalent statements, and 7.3.14 is also a characterization of the ultrafilters. In other words, we have the following.
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7.3.15 Theorem. Let X be a set. A filter U in X is an ultrafilter if and only if S each time that A1 , . . . , An ⊂ X are such that ni=1 Ai ∈ U, then Ai ∈ U for some i ∈ {1, . . . , n}. t u 7.3.16 Exercise. Show that every filter F in a set X is the intersection of all ultrafilters U that contain F. 7.3.17 Exercise. Let F be a filter in a set X such that there is a unique ultrafilter U which is finer than F. Show that F = U. 7.3.18 Exercise. Let X be a topological space and take A ⊂ X. (a) Show that x ∈ A◦ if and only if A ∈ F for every filter F that converges to x. (b) Show that x ∈ A if and only if there is at least one filter F that converges to x, such that A ∈ F. 7.3.19 Exercise. Let X be a topological space. Show that an ultrafilter U in X either converges or does not have cluster points. 7.3.20 Exercise. Let P be a set of subsets of a set X such that if P1 , P2 ∈ P, then P1 ∩ P2 , P1 ∪ P2 ∈ P. A Pfilter is a nonempty family F of nonempty sets in P such that (i) P1 , P2 ∈ F =⇒ P1 ∩ P2 ∈ F, and (ii) P1 ∈ F, P1 ⊂ P2 ∈ P =⇒ P2 ∈ F. A Pultrafilter is a maximal Pfilter. If X is a topological space, then an open filter in X is a Pfilter such that P is the set of open sets in X (i.e. its topology) and an open ultrafilter is a maximal open filter. (a) Show that every open filter is contained in an open ultrafilter. (b) Show that the next statements are equivalent: (1) U is un open ultrafilter. (2) If G is an open set in X such that G ∩ H 6= ∅ for all H ∈ U, then G ∈ U. (3) If G is an open set in X and G 6∈ U, then X − G ∈ U. 7.3.21 Exercise.
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7.4 Filters and functions
T (a) Let F be a filter in a set X such that A = F ∈F F 6= ∅. What relation do the filter F and the filter FA generated by A keep? (b) Does a filter F exist such that
7.4
T F ∈F
F = ∅?
Filters and functions
Let X and X 0 be sets and take a function f : X −→ X 0 . If B is a filter basis in X, then the images under f of the elements of B make up a filter basis in X 0 , since if we take B, B 0 ∈ B, then there is a set B 00 ∈ B such that B 00 ⊂ B ∩ B 0 . Hence f (B 00 ) ⊂ f (B) ∩ f (B 0 ). If F is a filter in X, then in general the image under f of its elements does not form a filter. For instance, if f is not surjective, then X 0 is not one of the images. However, given that any filter is a filter basis, the images of the elements of F under f also make up a filter basis. 7.4.1 Definition. Let f : X −→ X 0 be a function of sets, and let F be a filter in X. The filter generated by the images under f of the elements of F is called the image of F under f and is denoted simply by f (F). 7.4.2 Example. Let N have the usual order and take En = {k ∈ N  k ≥ n}. The sets En constitute a filter basis in N and the filter E generated by this filter basis is the filter whose elements are the sets in N with finite complement. A function N −→ X is precisely a sequence (xn ) in X and the image of E under this function is the elementary filter F(xn ) associated to this sequence. Take a filter basis B0 in the codomain X 0 of f . The inverse images under f of the elements of B 0 satisfy the axiom (FB), since f −1 (A0 ∩ B 0 ) = f −1 (A0 ) ∩ f −1 (B 0 ). If we wish that these inverse images form a filter basis, we need that for every B 0 ∈ B 0 , one has f −1 (B 0 ) 6= ∅. A sufficient condition for this is that f is surjective. If F 0 is now a filter in X 0 and f −1 (F 0 ) 6= ∅ for all F 0 ∈ F 0 , then {f −1 (F 0 )  F 0 ∈ F 0 } is a filter basis, but not no necessarily a filter. Namely, if F ⊃ f −1 (F 0 ), then there is not always a set F10 ∈ F 0 such that F = f −1 (F10 ). (Why? Exercise). 7.4.3 Definition. Let f : X −→ X 0 be a set function and let F 0 be a filter in X 0 . If the inverse images under f of the elements of F 0 form a filter basis in X, then the filter generated by this filter basis is called inverse image of F 0 under f and is denoted by f −1 (F 0 ).
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We have the following results. 7.4.4 Theorem. Let f : X −→ X 0 be a function of sets and let B 0 be a filter basis in X 0 . The filter F 0 generated by B 0 has an inverse image under f if and only if f −1 (B 0 ) 6= ∅ for each B 0 ∈ B 0 . t u 7.4.5 Theorem. Let f : X −→ X 0 be a function of sets. If F 0 = f (F) for some filter F in X, then there is an f −1 (F 0 ). Moreover, if f is surjective, then f −1 (F 0 ) exists for any F 0 . t u Take A ⊂ X and A0 ⊂ X 0 . If f : X −→ X 0 is a function, then one has that A ⊂ f −1 (f (A)) and that f (f −1 (A0 )) ⊂ A0 . In particular, if f is injective, then A = f −1 (f (A)) and if f is surjective, then f (f −1 (A0 )) = A0 . Hence we have the following. 7.4.6 Theorem. Let f : X −→ X 0 be a function of sets. (a) If F is a filter in X, then f −1 (f (F)) is a coarser filter than F and coincides with F if f is injective. (b) If F 0 is a filter in X 0 and f −1 (F 0 ) exists, then f (f −1 (F 0 )) is a finer filter than F 0 and coincides with F 0 if f is surjective. t u 7.4.7 Proposition. Let f : X −→ X 0 be a function of sets. If G is a finer filter than F in X, then f (G) is finer than f (F). Moreover, if F 0 is a finer filter than G 0 in X 0 , then f −1 (F 0 ) is finer than f −1 (G 0 ), in case that these exist. t u 7.4.8 Note. If one replaces the word “finer by “strictly finer in the previous theorem, then it does not hold anymore. (Exercise). 7.4.9 Note. If B is a basis of F, then f (B) = {f (B)  B ∈ B} is a basis of f (F). If on the other hand, B0 is a basis of F 0 , then f −1 (B 0 ) = {f −1 (B 0 )  B 0 ∈ B0 } is a basis of f −1 (F 0 ). Of course both statements only have sense if the corresponding filters do exist. 7.4.10 Proposition. Let f : X −→ Y be a surjective function of sets and let U be an ultrafilter in X, then f (U) is un ultrafilter en Y . Proof: Let U be an ultrafilter in X. Take A ⊂ Y . Then f −1 (Y − A) = X − f −1 (A) and hence, by 7.3.13, f −1 (A) or f −1 (Y − A) is an element of U, i.e. either A or Y − A is an element of f (U) and, again by 7.3.13, f (U) is an ultrafilter. t u
7.4 Filters and functions
151
7.4.11 Definition. Take A ⊂ X. Each filter G in A has an image in X under the inclusion A ,→ X, which is called continuation (or extension) of G to X. If X is a topological space and G is a filter in the subspace A of X such that its continuation to X converges in X, we shall simply say that G converges in X. This does not necessarily mean that G converges in A. However, if G converges in A, then also its continuation to X converges (to the same point of A). More concretely, one can solve the following exercise. 7.4.12 Exercise. Let X be topological space and take A ⊂ X and a filter G in A. Show (a) A point x ∈ A is a cluster point of G if and only if x is a cluster point of the continuation of G to X. (b) x ∈ A is a limit point of G if and only if x is a limit point of the continuation of G to X. If F is a filter in a set X, as we already saw, the filter i−1 (F) need not exist if i is the inclusion of A in X. Indeed, i−1 (F) exists if and only if F ∩ A 6= ∅ for all F ∈ F. In this case the filter i−1 (F) is called the trace of F in A. If A 6∈ F, but the trace of F in A exists, then the continuation of the trace is the supremum of F and the filter FA of all supersets of A. If, in particular, F is the neighborhood filter of a point x ∈ X, then from the definition of limit point of a set in a topological space, we obtain the following result, which generalizes Theorem 7.1.4 (a). 7.4.13 Theorem. Take A ⊂ X and x ∈ X. The following are equivalent: (a) x ∈ A. (b) The neighborhood filter of x has a trace in A. (c) There is a filter G in A such that G → x in X.
t u
Let X and Y be topological spaces and take a function f : X −→ Y and a point x0 ∈ X. In terms of filters, what does it mean that f is continuous in x0 ? By definition, this means that given a neighborhood of f (x0 ), there is a neighborhood of x0 whose image lies in the given neighborhood, namely Nf (x0 ) ⊂ f (Nx0 ). In other words, we have the following result. 7.4.14 Proposition. A map f : X −→ Y is continuous at x0 if and only if f (Nx0 ) → f (x0 ).
t u
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In case that this limit is unique, then we simply write (7.4.15)
lim f (x) = f (x0 ) .
x→x0
Hence, if Y is a Hausdorff space, then this equation characterizes the continuity of f at x0 . Now, if we have that f is continuous at x0 and F is a filter that converges to x0 , then F is finer than Nx0 , so that f (F) is finer than f (Nx0 ). This filter is finer than Nf (x0 ) . Hence we have the following result, which generalizes Theorem 7.1.4 (b). 7.4.16 Proposition. A map f : X −→ Y is continuous at x0 if and only if f (F) → f (x0 ) for every filter F such that F → x0 . t u In other words, continuous maps are precisely those maps which commute with limits, namely, such that f (lim F) = lim f (F). In particular we obtain Theorem 7.1.4 (b) as a consequence. We finish this section analyzing another fundamental concept in topology, which we already considered above 3.4.12 namely the concept of density. We start with a more general definition. 7.4.17 Definition. Let X be a topological space and take A, B ⊂ X. We say that A is dense with respect to B if B ⊂ A. We say that A is dense in B if A ⊂ B ⊂ A. 7.4.18 Lemma. If A is dense in X and Q is open in X, then A ∩ Q is dense in Q. Proof: We wish to show that Q ⊂ A ∩ Q. For this take x ∈ Q and a neighborhood U of x Since Q is open, U ∩ Q is a neighborhood of x, and since A is dense in X, then U ∩ Q ∩ A 6= ∅. Therefore x ∈ Q ∩ A. t u We have that if X is a Hausdorff space, then by 7.4.14, f : X −→ Y is continuous at x0 ∈ X if and only if limx→x0 f (x) = f (x0 ), namely, by definition, if and only if f (Nx0 ) → f (x0 ). If Nx0 induces a filter in A whose image under f A converges to y, then we may write that lim f (x) = y .
x→x0 x∈A
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7.5 Filters and products
In particular, if A = X − {x0 }, then we write lim f (x) = y .
x→x0 x6=x0
Assume that A is dense in X and that Y is a Hausdorff space. Then for any x0 ∈ X, the trace of Nx0 in A exists. If f : X −→ Y is continuous, then take g = f A , so that lim g(x) = f (x0 ) . x→x 0 x∈A
We have the next result. 7.4.19 Theorem. Let X and Y be topological spaces, Y Hausdorff, and take A ⊂ X dense. Let f, g : X −→ Y be continuous maps. If f A = gA , then f = g. t u 7.4.20 Exercise. Prove the previous theorem directly using explicitly the facts that A = X y and that Y is Hausdorff. 7.4.21 Exercise. Let f : X −→ Y be a continuous map and let D ⊂ X be a dense subset. Show that f (D) is dense in f (X) ⊂ Y . In particular, if f is surjective, then f (D) is dense in Y . (Hint: Use 2.5.8 (d).)
7.5
Filters and products
Filters are a good tool for studying the product topology. En this section we shall analyze the behavior of filters with respect to products. We start with the next result, which characterizes the filter convergence in a product. Q 7.5.1 Theorem. Take a product of topological spaces X = λ∈Λ Xλ and let F be a filter in X. Then F → x = (xλ ) if and only if, for all λ ∈ Λ, each image filter pλ (F) → xλ , where pλ : X −→ Xλ is the projection. Proof: Since pλ is continuous, if F converges to x, then pλ (F) converges to pλ (x). Conversely, assume that pλ (F) → xλ for all λ, and let Qλ be an open neigh−1 borhood of xλ in Xλ . Therefore Qλ ∈ pλ (F) and thus p−1 λ (Qλ ) ∈ pλ pλ (F) ⊂ F. Consequently F contiains also finite intersections of elements of the form p−1 λ (Qλ ), Q namely sets of the form λ∈Λ Qλ , where Qλ = Xλ , except for finitely many λ ∈ Λ. Since these sets form a neighborhood basis of x = (xλ ) in X, we have that Nx ⊂ F, i.e. F → x. t u
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7.5.2 Example. Let RR be the product of copies of R, one for each t ∈ R. In other words, RR = {f : R −→ R} with the product topology. Then fn → f in RR if and only if fn (t) → f (t) for all t ∈ R. In other words, one has function convergence in the product topology if and only if one has pointwise convergence. Theorem 7.5.1 provides a characterization of the product topology by stating which filters are convergent. We ask now generally about the existence of filters Q in the cartesian product λ∈Λ Xλ of sets Xλ with given projections. We already suggested in Example 7.1.8(d) how to construct a filter basis in a product, given a filter basis in each factor. Thus let Fλ be a filter in a set Xλ , λ ∈ Λ. Take κ ∈ Λ and Fκ ∈ Fκ . Then Q −1 pκ (Fκ ) = λ∈Λ Fλ , where Fλ = Fκ if λ = κ and Fλ = Xλ if λ 6= κ. The supremum Q of {p−1 λ (Fλ )  λ ∈ Λ} exists and is generated, precisely, by { λ∈Λ Fλ  Fλ ∈ Fλ , and Fλ = Xλ , except for finitely many λ ∈ Λ}. 7.5.3 Definition. Let Λ 6= ∅ and Xλ be a nonempty set for each λ ∈ Λ. The filter F generated by the filter basis Y Fλ  Fλ ∈ Fλ , and Fλ = Xλ , except for finitely many λ ∈ Λ} { λ∈Λ
is called product of the filters Fλ and is denoted by pλ (F) = Fλ . Compare with 7.1.8(d).
Q
λ∈Λ Fλ
and is such that
Take now nonempty topological spaces Xλ , λ ∈ Λ, and let Fλ be a filter in Xλ Q Q for each λ. If λ∈Λ Fλ has two different limits in λ∈Λ Xλ , then Fλ also has two different limits in Xλ para some λ. Conversely, if for some λ there is a filter Fλ which has two different limits in Xλ , then (if for each λ, Xλ 6= ∅) there is a filter Q F with two different limits in X = λ∈Λ Xλ . Thus we have, as a first application of filters to products of topological spaces, the following. 7.5.4 Theorem. Let {Xλ  λ ∈ Λ} be a nonempty family of nonempty topological Q spaces and take X = λ∈Λ Xλ . Then X is Hausdorff if and only if Xλ is Hausdorff for every λ ∈ Λ. t u Take Λ 6= ∅ and for each λ ∈ Λ, take a nonempty set Xλ . Let U be an ultrafilter Q in cartesian the product X = λ∈Λ Xλ . Take Uλ = pλ (U) and let Fλ be a filter in Q Xλ , which is finer than Uλ . Then F = λ∈Λ Fλ is a finer filter than U. But since U is an ultrafilter, F = U. Due to the fact that pλ (F) = Fλ and pλ (U) = Uλ , we conclude that Fλ = Uλ . Therefore Uλ is also an ultrafilter. Q Conversely, if Uλ is an ultrafilter for each λ ∈ Λ, then take U = λ∈Λ Uλ and let F be a finer filter in X than U. Hence pλ (F) is finer than pλ (U) = Uλ . Hence pλ (F) = Uλ and consequently F = U. Thus we have tha next result.
7.6 Nets
155
7.5.5 Theorem. Take a nonempty family {Xλ  λ ∈ Λ} of nonempty sets. Then Q U = λ∈Λ Uλ is an ultrafilter if and only if Uλ is un ultrafilter for each λ ∈ Λ. t u 7.5.6 Exercise. Let {Xλ  λ ∈ Λ} be a nonempty family of nonempty topological spaces and for each λ ∈ Λ, let Fλ be a filter in Xλ . Let F the product filter in Q X = λ∈Λ . Prove (a) x = (xλ ) ∈ X is a cluster point of F if and only if xλ ∈ Xλ is a cluster point of Fλ for all λ. (b) F → x = (xλ ) ∈ X if and only if Fλ → xλ ∈ Xλ for all λ. If pλ = projXλ : X −→ Xλ , then conclude that for an arbitrary filter F in X the following hold: (c) If xλ ∈ Xλ is a cluster point of Fλ = pλ (F), then x = (xλ ) ∈ X is a cluster point of F, (d) If Fλ = pλ (F) → xλ ∈ Xλ , then F → x = (xλ ) ∈ X.
7.6
Nets
When a topological space is firstcountable, we have shown that sequenceconvergence characterizes its topology. Nets are a generalization of sequences and their convergence can be used to characterize the topology of arbitrary spaces in a very similar way as sequenceconvergence does in the special case of firstcountable spaces. In contrast to filters, nets behave in a very similar way to sequences and therefore are easy to use. In some sense, nets and filters are equivalent: Given a net, there is a filter whose convergence corresponds to that of the net. Conversely, given a filter, one can associate a net, with similar convergence. In this section we shall prove the most important results about nets and topological spaces using the relation ship between nets and filters. 7.6.1 Definition. A set M is called a directed set if it has a preorder relation ≤ that satisfies the first two order axioms (OR1) (the relation is reflexive) and (OR2) (the relation is transitive) of 7.3.1, and the following additional axiom: (DS) For any a, b ∈ M there exists c ∈ M such that a ≤ c and b ≤ c. 7.6.2 Examples. The following are directed sets:
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(a) M = R, Z or N with the usual order relation ≤. (b) M = P(X) the power set of a set X with the preorder relation ≤ = ⊇ (or ≤ = ⊆). (c) M = Nx , a local neighborhood basis at a point x in a topological space X, with the preorder relation ≤ = ⊃ (or ≤ = ⊂). The relation ≤ is called a direction in M . In order to study convergence in an adequate set up in a topological space, the directed set of Example (c) will be needed. The next definition generalizes sequences. 7.6.3 Definition. Let X be a set and M be a directed set. An M net, or simply a net in X is a function M −→ X. For each a ∈ M we denote by xa ∈ X the image of a in X and we represent the net by (xa )a∈M , or simply by (xa ). Let M 0 be a directed set and ϕ : M 0 −→ M an increasing and cofinal function, that is, it satisfies a0 ≤ b0 =⇒ ϕ(a0 ) ≤ ϕ(b0 ) (ϕ is increasing) and given a ∈ M, there is some a0 ∈ M 0 such that a ≤ ϕ(a0 ) (ϕ is cofinal). We define a subnet of (xa )a∈M as the M 0 net (xϕ(a0 ) )a0 ∈M 0 . In analogy to sequences, one can define the tail of (xa ) as the set (xa )b = (xa )b≤a , for some b ∈ M . Since given b1 , b2 ∈ M , there exists b ∈ M such that b1 , b2 ≤ b, one has that {xa }b ⊂ {xa }b1 ∩ {xa }b2 . Hence, the set of tails of a net is a filter basis which generates a filter F(xa ) , which is called filter generated by the net (xa ). Since netconvergence is modeled after sequenceconvergence, the following definitions should be clear. 7.6.4 Definition. Let (xa )a∈M be a net in X and let A ⊂ X. We say that the net (xa ) lies frequently in A if for every a ∈ M there exists b ∈ M , a ≤ b such that xb ∈ A and that it lies finally in A if there exists a ∈ M such that for every b ∈ M , a ≤ b, xb ∈ A. Assume now that X is a topological space. A net (xa ) converges to a point x ∈ X (in symbols, (xa ) → x or simply xa → x) if for any neighborhood V of x
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the net lies finally in V . Then one defines x as a limit point of the net (in symbols, x ∈ lim xa ). One says that y is an cluster point (or accumulation point) of a net (xa ) if for any neighborhood W of y the net lies frequently in W . If, for instance, X is a discrete space, then a net (xa ) converges to x ∈ X if and only if (xa ) lies finally in {x}, namely if and only if there is b such that if b ≤ a, then xa = x. On the other hand, if X is indiscrete (with more than one point), then every net (xa ) converges to any point in X, that is, lim xa = X. Thus, a net may have several limit points. 7.6.5 Exercise. Prove that if a net (xa ) in X lies finally in a set V ⊂ A, then it also lies frequently in it. Conclude that any limit point of a net in a topological space is also an cluster point of the net. 7.6.6 Exercise. Prove that if a net (xa ) in a topological space X is finally constant (say finally x), then (xa ) converges (namely xa → x). 7.6.7 Examples. (a) Let X be a topological space, x ∈ X, and let M be any local neighborhood basis in X at the point x. As in 7.6.2(c), we give M a relation by U ≤ V if and only if V ⊂ U , that makes it to a directed set. Take a point xU ∈ U for every U ∈ M . Then (xU ) is a net in X such that xU → x. Namely, given any neighborhhod W of x in X, there is some U ∈ M such that U ⊂ W . Therefore, if U ≤ V in M , then V ⊂ W , and so xV ∈ W . Thus (xU ) lies finally in the given neighborhood W . (b) Since the set N of the natural numbers with their standard order relation is a directed set, every sequence (xn ) is a net. Clearly, (xn ) converges to x as a sequence if and only if it does converge to x as a net. Observe that every subsequence of (xn ) is a subnet of the net (xn ); however, the converse is not true. One may give a subnet of (xn ) that is not a subsequence. (c) Recall that a partition of a closed interval [a, b] is a finite sequence P = {t0 = a < t1 < t2 < · · · < tk = b} and that another partition Q is a refinement if P ⊂ Q. We write this fact by P ≤ Q. Thus the set P of all partitions of [a, b] is a directed set. Given any realvalued function f on [a, b], we can define a net L(f ) : P −→ R by letting L(f )(P ) be the lower Riemann sum of f over the partition P and U (f ) : P −→ R by letting U (f )(P ) be the upper Riemann sum of f over the partition P (see [14, 6.1]). That both of these two nets converge to the same value c means that f is integrable, and Z b f (t)dt = c . a
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This example is what led Moore and Smith to the concept of a net (see [13]). 7.6.8 Exercise. Let X be a metric space and take x0 ∈ X. Prove that the complement of this point M = X − x0 becomes a directed set when preordered by the relation x < x0 if and only if d(x0 , x0 ) < d(x, x0 ). Take f : X −→ Y , where Y is another metric space. The restriction f M : M −→ Y defines a net in Y . Prove that this net converges to y0 ∈ Y if and only if limx→x0 f (x) = z0 in the sense of elementary calculus. 7.6.9 Exercise. Prove that if a net (xa ) in a topological space X converges to x, then every subnet of (xa ) converges to x. The following result relates net convergence and filter convergence. 7.6.10 Proposition. Let X be a topological space and let (xa ) be a net in X. Then xa → x if and only if F(xa ) → x. Proof: If xa → x, then for each neighborhood V ∈ Nx there is a tail of the net (xa ) which is contained in V . Therefore V ∈ F(xa ) , i.e. Nx ⊂ F(xa ) . Thus F(xa ) → x. Conversely, if F(xa ) → x, namely, if Nx ⊂ F(xa ) and V ∈ Nx , then V contains a tail of the net, that is, the net lies finally in V . Hence xa → x. t u Thus, since convergence for nets and for sequences are consistent concepts, so are the concepts of cluster point too. Namely, we have the next result, which generalizes Theorem 7.2.7 and whose proof is similar to the previous proof. 7.6.11 Proposition. Let X be a topological space and take a net (xa ) in X. Then y is a cluster point of the net (xa ) if and only of if y is a cluster point of the filter F(xa ) generated by the net. t u From 7.2.13, the next assertion is immediate. 7.6.12 Proposition. Let X be a Hausdorff space. Then every net (xa ) in X converges to at most one point x ∈ X. We can use nets to characterize the closure of a set in a topological space and with it, to characterize the topology. 7.6.13 Theorem. Let X be a topological space and take A ⊂ X. The following are equivalent:
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(a) x ∈ A. (b) There is a net (xa ) in A which converges to x. (c) There is a net (xa ) in A which has x as cluster point. Proof: (a) =⇒ (b) For each V ∈ Nx , V ∩ A 6= ∅ (since x ∈ A). Let M = Nx be the directed set of Example 7.6.2(c) and let M → X be a net such that xV ∈ V ∩ A for each V ∈ M . Thus the net (xV ) lies in A and for each V ∈ Nx one has that xW ∈ V if V ≤ W , that is, if V ⊃ W . Therefore xV → x. (b) =⇒ (c) Every limit point of a net is a cluster point, since if a net lies finally in some neighborhood, then it also lies frequently in it (see 7.6.5). (c) =⇒ (a) Take V ∈ Nx . Since x is a cluster point of a net (xa ) in A, the net lies frequently in V , i.e. for each a there is a b with a ≤ b such that xb ∈ V . Therefore, since xb ∈ A, we have A ∩ V 6= ∅ so that x ∈ A. t u As a consequence, we have the following characterization of the topology of a space using net convergence, which generalizes 7.1.4(a) to spaces that not necessarily are firstcountable. 7.6.14 Proposition. Let X be a topological space and take A ⊂ X. The following are equivalent: (a) A is closed. (b) For every net (xa ) in A such that y ∈ X is a cluster point of it, one has y ∈ A. (c) For every net (xa ) in A such that x ∈ X is a limit point of it, namely xa → x, one has x ∈ A. Proof: (a) =⇒ (b) By Proposition 7.6.13, y ∈ A, so that, by (a), y ∈ A. (b) =⇒ (c) Every limit point of a net is a cluster point. Thus x ∈ A. (c) =⇒ (a) Take x ∈ A. By 7.6.13, there is a net (xa ) in A which converges to x. Hence, by (c), x ∈ A, so that A ⊂ A, namely, A is closed. t u Let f : X −→ Y be a map between topological spaces and take a net (xa )a∈M in X. Then there is a net (f (xa ))a∈M in Y , which we call image of the net (xa )a∈M under f . Convergence of nets allows to characterize continuity. We have the next result.
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7.6.15 Proposition. A map between topological spaces f : X −→ Y is continuous in a point x0 ∈ X if and only if, for every net which converges to x0 , xa → x0 , the image net converges to f (x0 ), f (xa ) → f (x0 ). Proof: Assume first that xa → x0 and take V ∈ NfY(x0 ) . Since f is continuous at x0 , there is a neighborhood U ∈ NxX0 such that f (U ) ⊂ V . Since the net converges, it lies finally in U . Thus its image (f (xa )) lies finally in f (U ) and hence in V . Therefore f (xa ) → f (x0 ). Conversely, if f is not continuous at x0 , there is a neighborhood V ∈ NfY(x0 ) such that for all neighborhoods U ∈ NxX0 , f (U ) 6⊂ V , namely, f (U ) ∩ (Y − V ) 6= ∅. Since M = NxX0 is a directed set, then as we said in Example 7.6.2(c), we may define an M net in X, (xU ) such that for each U ∈ M , the element xU ∈ U is such that f (xU ) 6∈ V . By construction, xU → x0 , but f (xU ) 6→ f (x0 ), since this image net does not lie finally in V . t u We put in one theorem, which generalizes Theorem 7.1.4, the previous results. 7.6.16 Theorem. Let X be a topological space. Then (a) A ⊂ X is closed if and only if for every net (xa ) in A such that, if xa → x0 in X, then x0 ∈ A. (b) f : X −→ Y is continuous at x0 ∈ X if and only if for every net (xa ) in X such that xa → x0 , the image net f (xa ) → f (x0 ). We thus have that the common techniques of sequences used for checking topological properties of firstcountable spaces can be used for arbitrary spaces, provided that one takes nets instead of sequences. 7.6.17 Exercise. Let F be a filter in a set X. (a) Put MF = {(x, F )  F ∈ F and x ∈ F }. Prove that with the order relation (x, F ) ≤ (y, G) ⇐⇒ F ⊃ G, MF is a directed set. The MF net in X, MF −→ X such that (x, F ) 7→ x is called the net determined by the filter F. (b) Prove that the filter determined by the MF net such that x(x,F ) = x is precisely F. (c) Let X be a topological space and take a filter F in X. Show that F → x0 if and only if the net determined by F converges to x0 .
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(d) Let M be an arbitrary directed set and take an M net (xa ) in X. There is a function M −→ MF(xa ) given by a 7→ (xa , (xb )a ). Show that this function preserves the order. Moreover, the net (x, F ) 7→ x extends the given net (xa ), i.e. the composition M −→ MF −→ X (x, F ) 7→ x is the given net. Show that this last net converges to x0 if and only if the given net converges to x0 (e) Let F be a filter in X and let MF −→ X, (x, F ) 7→ x, be the net determined by F. Prove that the filter F converges to x0 if and only if the filter F(x(x,F ) ) converges to x0 . The previous exercise shows that from the point of view of convergence, net theory and filter theory are equivalent. 7.6.18 Exercise. Let X be a topological space. Show that if a net (xa ) has a cluster point x, then the net has a subnet (yb ) which converges to x.
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Chapter 8
Compactness
Compactness is one of the most important conditions needed to prove fundamental results in topology and analysis. En this chapter we shall study basic concepts related to compactness. We start analyzing several conditions equivalent to compactness and giving simple and deep applications of the concept. We shall study two fundamental theorems on compactness, namely the Heine– Borel–Lebesgue Theorem and the Bolzano–Weierstrass Theorem, as well as their generalizations to metric spaces and, more generally, to firstcountable spaces. Further on we shall study the concept of compactification, that refers to how one can densely embed noncompact spaces in compact spaces. This is a very important issue in several areas such as algebraic geometry and topology. We shall analyze here the onepoint (or Alexandroff) compactification. In the next ˇ chapter we shall study the Stone–Cech compactification. We shall finish the chapter we some applications of compactness to construct other classes of topological spaces which have nice properties and are useful for homotopy theory. These are the categories of compactly generated spaces and of kspaces. We shall show to alter the topology of a given space to convert it into one of each of these classes. These changes of topology produce finer topologies in each case, however, the usual algebraic invariants studied in algebraic topology remain unchanged.
8.1
Compact sets
We start by briefly recalling the two main results on compactness in Rn . For the time being, we shall omit the proof, but by the end of the chapter we shall give it. 8.1.1 Theorem. (Heine–Borel–Lebesgue) In Rn , a set A is closed and bounded if and only if every open cover of A contains a finite subcover. 8.1.2 Theorem. (Bolzano–Weierstrass) In Rn , a set A is closed and bounded if and only if every sequence in A has an cluster point in A. 163
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Further on (8.2.12 and 8.2.13) we shall give the proof of these results, after we study the general theory of compactness. 8.1.3 Definition. Let X be topological space and A ⊂ X. We say that a family S C = {Uλ ⊂ X  λ ∈ Λ} is a cover of A in X if the union λ∈Λ Uλ ⊃ A. We say that the cover C is open, resp. closed, if Uλ is open, resp. closed, in X for all λ ∈ Λ. If C 0 ⊂ C and C 0 is again a cover of A, then we say that C 0 is a subcover of C. In particular, we say that C 0 is finite, resp. countable, if as a set C 0 is finite, resp. countable. 8.1.4 Theorem. Let X be a topological space. The following statements are equivalent: (C) Every open cover of X contains a finite subcover. (C0 ) Every family of closed sets in X, whose intersection is empty, contains a finite subfamily whose intersection is empty. (C00 ) Given a family of closed sets in X such that any finite subfamily has an nonempty intersection, then the given family has nonempty intersection. (C000 ) Each filter in X has, at least, one cluster point. (Civ ) Every ultrafilter in X converges. Proof: (C)⇐⇒(C0 ) is clear since {Uλ  λ ∈ Λ} is an open cover if and only if {X − Uλ  λ ∈ Λ} is a family of closed sets with empty intersection. (C0 )⇐⇒(C00 ) is obvious. (C00 )=⇒(C000 ) Let F be a filter in X. Each finite collection of closed sets F ∈ F T has nonempty intersection, therefore, by (C00 ), I = {F  F ∈ F, F is closed } = 6 ∅. Take x ∈ I, hence x is a cluster point of F, since if V ∈ Nx and G ∈ F, then x ∈ G ∈ F and thus V ∩ G 6= ∅. T (C000 )=⇒(C00 ) Let G be a family of closed sets such that ki=1 Gi 6= ∅, Gi ∈ G, T k ∈ N, and let F = {F ⊂ X  F ⊃ ki=1 Gi , Gi ∈ G, k ∈ N}. F is clearly a filter and by (C000 ), it has a cluster point x. Therefore every neighborhood of x meets Tk T {G  G ∈ G}. i=1 Gi , Gi ∈ G, k ∈ N. Hence x ∈ G for each G ∈ G, and thus x ∈ (C000 )⇐⇒(Civ ) is clear.
t u
8.1.5 Definition. A topological space X that satisfies one, and hence all, conditions (C),...,(Civ ), is said to be compact. A subset A of a topological space X is compact if as a subspace of X with the relative topology is a compact space.
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8.1.6 Proposition. Let X be a compact space. If a filter F in X has a unique cluster point x ∈ X, then F converges to x. Proof: If F did not converge to x, then there would be an open neighborhood V of x such that V 6∈ F. Hence, by 7.1.25, there would be a filter G finer than F such that X − V ∈ G. Since X is compact, G would have a cluster point y ∈ X. Since y belongs to X − V , y and x would be different. But y would be also a cluster point of F, contardicting the uniqueness of the cluster point. t u 8.1.7 Examples. (a) Every indiscrete space is compact. (b) A discrete space is compact if and only if it is finite. (c) Every finite space is compact. (d) The Cantor set defined in 2.2.7 is compact. (e) Every space X with the cofinite topology is compact. Namely, if {Uλ } is an open cover, then for any λ1 , the set X − Uλ1 is finite. Hence we can choose a finite number of open sets Uλ2 , . . . , Uλk of the cover, whose union contains this finite set. Then {Uλ1 , Uλ2 , . . . , Uλk }is a finite subcover of X 8.1.8 Exercise. Prove that the Cantor set C defined in 2.2.7 is homeomorphic to a product of a countable family of discrete spaces, each having two elements. This shows that the product of discrete spaces need not be discrete. (Hint: Each P xi point x ∈ I can be ternarily expressed as a sum , where xi ∈ {0, 1, 2}. Thus, 3i to each x corresponds an expression (x1 , x2 , x3 , . . . ). These expressions are unique up to the fact that each number, except 1, whose ternary expression ends with a sequence of 2s, can be expressed by one which ends with a sequence of 0s. For instance, 13 can be expressed by the sequence (1, 0, 0, 0, . . . ) or as (0, 2, 2, 2, . . . ). Hence the Cantor set C consists precisely of the points x whose ternary expression does not contain 1s. If D = {0, 2} has the discrete topology, then the function Q C −→ N D given by x 7→ (x1 , x2 , x3 , . . . ) determines a homeomorphism. The set C is not discrete, since in a discrete set the only convergent sequences are those which are eventually constant, while in C the sequence { 31n } converges to 0.) 8.1.9 Exercise. Using the fact that every point in the Cantor set has a unique expression as a sequence x1 , x2 , x3 , . . . , where xi = 0, 2, show that it is an uncountable set. (Hint: Emulate the proof that the set of real numbers is uncountable.)
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8.1.10 Exercise. Let X be a topological space. Show that X is compact if and only if every net {xa } in X has a cluster point. Consequently X is compact if and only if every net in X has a convergent subnet. (See 7.6.18.) 8.1.11 Proposition. Let X be a topological space and take a subset A ⊂ X. Then A, with the relative topology, is compact if and only if every open cover {Uλ } of A in X contains a finite subcover. Proof: Let A be compact as a space with the relative topology and let {Uλ } be an open cover of A in X. {Uλ ∩ A} is a cover of A by sets which are open in A. Since A is compact, there is a finite number of open sets U1 ∩ A, . . . , Uk ∩ A in this cover such that A = U1 ∩ A ∪ · · · ∪ Uk ∩ A. Hence A ⊂ U1 ∪ · · · ∪ Uk . S Conversely let {Vλ } be a family of open sets of A such that A = Vλ . Since A has the relative topology, there are open sets Uλ in X such that Vλ = Uλ ∩ A. S Therefore A ⊂ Uλ . By assumption, we can take a finite number of such open sets U1 , . . . , Uk such that A ⊂ U1 ∪ · · · ∪ Uk . Consequently A = V1 ∪ · · · ∪ Vk . t u The usual way of defining the concept of compact set is as in the statement of the previous proposition. This way it might seem that compactness would be a property that depends of the way in which the set is embedded in the space. However the statement of the proposition says that the compactness concept is inherent to the set seen as a topological space, rather than to the way it lies inside the larger space. 8.1.12 Remark. Not every subspace of a compact space is compact. For example, using the Heine–Borel–Lebesgue theorem for R, (0, 1) is not compact. However [0, 1] is compact. To see directly that the open interval (0, 1) is not compact, consider the open cover {( n1 , 1 − n1 )}, n ∈ N. This cover does not contain a finite subcover. 8.1.13 Theorem. Let X be a compact topological space and take A ⊂ X. If A is closed, then A is compact. Proof: Let {Fλ } be a family of closed sets in A whose intersection is empty. Since A is closed in X, the sets Fλ are closed in X for all λ. Since X is compact, there exists a finite subfamily of the given family, whose intersection is empty. Hence A is compact. t u 8.1.14 Exercise. Provide proofs of Theorem 8.1.13 using other conditions for compactness.
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8.1.15 Theorem. Let X be a Hausdorff space and take A ⊂ X. If A is compact, then A is closed. Proof: Assume A 6= ∅ and take x ∈ A. The neighborhood filter Nx of x in X induces a filter F in A and since A is compact, F has a cluster point a ∈ A. This point a is also a cluster point of the continuation F 0 = i(F) = ii−1 (Nx ) of F to X, which is finer than Nx , where i : A ,→ X is the inclusion map. Hence F 0 → x and since X is a Hausdorff space, x is the only cluster point of F 0 . Thus x = a. Therefore x ∈ A and so A is closed in X. t u Theorem 8.1.15 and Proposition 8.1.13 can be written together as follows. 8.1.16 Corollary. Let X be a compact Hausdorff space and take A ⊂ X. Then A is compact if and only if A is closed. t u 8.1.17 Exercise. Let X be a Hausdorff space and take a compact subset A ⊂ X. If x ∈ X − A, show that there are disjoint open sets U and V in X such that x ∈ U and A ⊂ V . Observe that this assertion generalizes 8.1.15. 8.1.18 Theorem. Let X be a topological space. Every finite union compact sets Ai in X is a compact set.
Sn
i=1 Ai
of
S Proof: Take open cover U = {Uj }j∈J of ni=1 Ai . Then U is an open cover of Ai for all i = 1, . . . , n. Since each Ai is compact, there are Ui,1 , . . . , Ui,mi ∈ U that cover Ai . Hence the finite family U1,1 , . . . , U1,m1 , . . . , Un,1 , . . . , Un,mn ∈ U covers
Sn
i=1 Ai .
t u
8.1.19 Theorem. Let X be a Hausdorff space. Every nonempty intersection T i∈I Ai of compact sets Ai in X is a compact set. T Proof: Since X is a Hausdorff space, each Ai is closed and thus i∈I Ai is closed. Since this intersection is also a closed subset of each Ai , which is compact, then the intersection is compact. t u There are spaces X that are not Hausdorff spaces, where we can find compact sets whose intersection is not compact. Namely, we have the following examples.
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8.1.20 Example. Take X = N with the topology A = {∅, N, A  A ⊂ 2N}. The set Ck = {k} ∪ 2N is compact if k is odd. However Ck ∩ Cl = 2N, where k, l are odd, k 6= l, is not compact, since it is discrete and infinite. 8.1.21 Example. Let Y = {0, 1} be furnished with the indiscrete topology and take X = R × Y with the product topology. The sets A = (0, 3) × {0} ∪ {0, 3} × {1} y
B = (1, 2) × {0} ∪ {1, 2} × {1}
are compact. For any a < b ∈ R, the map q : [a, b] −→ (a, b) × {0} ∪ {a, b} × {1} , given by q(x) = (x, 0) if x 6= a, b and by q(a) = (a, 1), q(b) = (b, 1), is continuous and surjective. Therefore, by 8.1.22 below, since the interval [a, b] ⊂ R is compact, (a, b) × {0} ∪ {a, b} × {1} is also compact. Hence A and B ar compact. However, we have that the intersection A ∩ B = (1, 2) × {0} is not compact. The following result provides us with one of the most important properties of continuity related to compactness. 8.1.22 Theorem. Let X be a compact space and let f : X −→ Y be a continuous map. Then the image of X under f , f (X), is compact. Proof: Let C be an open cover of f (X) in Y and consider the family f −1 (C) = {f −1 (U )  U ∈ C}. Since f is continuous, f −1 (C) is an open cover of X. Since X is compact, there are finitely many open sets U1 , . . . , Uk such that X = f −1 (U1 ) ∪ · · · ∪ f −1 (Uk ). Therefore f (X) ⊂ U1 ∪ · · · ∪ Uk . t u 8.1.23 Exercise. Let X be a nonempty compact Hausdorff space and let f : X −→ X be continuous. Prove that there exists a nonempty closed set A ⊂ X such that f (A) = A. 8.1.24 Definition. Let X be a topological space and take A ⊂ X. We say that A is relatively compact if the closure A is compact. 8.1.25 Examples. (a) By the Heine–Borel–Lebesgue Theorem, any bounded set in a Euclidean space is relatively compact. Conversely any relatively compact set in a Euclidean space is bounded.
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(b) If X is a compact space, then every subset of X is relatively compact. The next example due to Ra´ ul P´erez shows that not every compact set is relatively compact. Of course, this can only happen inside a nonHausdorff space. 8.1.26 Example. Take the set of irrational numbers of the unit interval together with 1, namely X = I ∩ (R − Q) ∪ {1} and take the surjective map q : (0, 1] −→ X given by ½ x if x ∈ I ∩ (R − Q) q(x) = 1 if x ∈ I ∩ Q . Give X the identification topology. The topological space X is neither Hausdorff nor compact. To see this take a sequence 1 > x1 > x2 > x3 > · · · of positive irrationals that tends to zero (in the usual topology). Then the family {Un } given S by Un = q( k≥n (xk+1 , xk ) ∪ (xk , 1]) ⊂ X is an open cover of X that does not contain a finite subcover. Namely, take 0 < a < b < 1 and A = q[a, b] ⊂ X. Since [a, b] is compact (Heine–Borel Theorem) and q is continuous, then A is a compact set. However its closure A coincides with X, which is not compact. To see this take any nonempty open subset U of X that contains 1 as an element and since q −1 (U ) is an open subset that contains I ∩ Q, then it also contains irrational numbers in any interval, in particular in [a, b]. Thus A ∩ U 6= ∅. Therefore A is not compact, i.e. A is not relatively compact. The previous example shows the convenience of working with Hausdorff spaces, as many authors assume of their compact spaces. In particular, if X is Hausdorff, then every compact set A ⊂ X is relatively compact, since it is already closed (because X is Hausdorff), and so A = A. By the way, that example also shows that identification spaces, even of the nicest spaces such as an interval, can be quite nasty. Assume now that A is a relatively compact subset of a topological space X. Therefore any filter in A is such that its extension to the closure A has a cluster point. Therefore its extension to all of X has also a cluster point, since a filter F in X is the extension of a filter in A if and only if A ∈ F. Then we have the following. 8.1.27 Theorem. Let X be a topological space and let A ⊂ X be relatively compact. If F is a filter in X that contains A, then F has a cluster point. t u In what follows, we shall prepare the proof of the Heine–Borel–Lebesgue Theorem.
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8.1.28 Definition. If X is a metric space, we say that a subset A ⊂ X is bounded if A ⊂ Bn (x) for some point x ∈ X and some natural number n. 8.1.29 Exercise. Prove that if X is a metric space, then a subset A ⊂ X is bounded if and only if for any point x ∈ X there exists n ∈ N such that A ⊂ Bn (x). 8.1.30 Lemma. Every compact set A in a metric space X is closed and bounded. Proof: Since X is Hausdorff, then A is closed. Given any point x0 ∈ X, the family of open balls {Bn◦ (x0 )  n ∈ N} is clearly an open cover of X and, in particular, of A. Since A is compact, we may find a finite subcover. Therefore A ⊂ Bn◦ (x0 ) for some n ∈ N, since each open ball in the cover is contained in the next one with respect to the order of the naturals. Consequently A is bounded. t u 8.1.31 Note. Remember (1.3.6) that given a metric space X with metric d, we may find a bounded metric d0 such that it determines the same topology on X. Therefore, we have that any metrizable space admits a bounded metric and therein, every set, including X itself, is bounded. This shows that the inverse of Lemma 8.1.30 is false in general. If X is a compact space and f : X −→ R is continuous, then Theorem 8.1.22 claims that the image of f (X) ⊂ R is a compact set, and by the previous lemma, it is closed and bounded. We have therefore the following consequence. 8.1.32 Corollary. Any real continuous function f : X −→ R defined on a compact space X reaches its maximum and its minimum. In other words, there are points x0 , x1 ∈ X such that f (x0 ) ≤ f (x) ≤ f (x1 ) for all x ∈ X. t u 8.1.33 Exercise. Show that every metric d in a compact space X is bounded, i.e. there exists K > 0 such that d(x, y) ≤ K for all x, y ∈ X. 8.1.34 Theorem. Let X and Y be topological spaces and let f : X −→ Y be continuous. If X is compact and Y is Hausdorff, then f is a closed map. Moreover, if f is surjective, then f an identification. Proof: Let A ⊂ X be closed. Since X is compact, A is compact, and since f is continuous, f (A) is compact. Moreover, since Y is Hausdorff, f (A) is closed. Therfore f is closed. If we now also assume that f is surjective and B ⊂ Y is such that f −1 (B) is closed, then, since f closed, the image B = f f −1 (B) is closed. This proves that f is an identification. t u
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The following is a very useful result, which clearly follows from the previous theorem. 8.1.35 Corollary. Let X be a compact space and Y a Hausdorff space, and let f : X −→ Y be a continuous bijective map. Then f is a homeomorphism. t u 8.1.36 Exercise. Using geometric arguments, and not necessarily explicit formulas, show that there are identifications B2 −→ S1 , S2 −→ B2 , S2 −→ S1 , S2 −→ S1 × S1 , and S1 × S1 −→ S2 . 8.1.37 Remark. The combination of the properties of being compact and Hausdorff is particularly interesting. Namely, let X be a compact Hausdorff space. If X 0 denotes another space with the same underlying set and a coarser topology, then the identity map f = id : X −→ X 0 is continuous. Hence, if A ⊂ X is closed and thus compact, then f (A) ⊂ X 0 is compact. If X 0 were Hausdorff, then f (A) would be closed, so that f would be a closed map. Then the topology of X 0 would be finer than that of X and so X = X 0 . We have shown that the topology of X is the coarsest that, being compact, is Hausdorff. Conversely, if now X 0 were again a topological space with the same underlying set as X, but with a finer topology, el espacio con el then f = id : X 0 −→ X would be continuous. Therefore, f would be a bijective map from a compact space to a Hausdorff space. Thus it would be a homeomorphism. Again the topology of X would be the same as that of X 0 . This shows that the topology of X is the finest that, being Hausdorff, is compact. The previous remark proves the next. 8.1.38 Theorem. The topology of a compact Hausdorff space X is the finest that makes it compact and the coarsest that makes it Hausdorff. This makes the topology of a compact Hausdorff space in some sense unique. t u 8.1.39 Definition. We say that a topological space X has the maximal compact if X is compact and with any strictly finer topology it is not compact any more. The converse of Theorem 8.1.38 is not true, that is, a minimal Hausdorff topology, does not have to be compact, nor a maximal compact topology has to be Hausdorff. To see an example, the one may read [17, Examples 99 and 100]. 8.1.40 Exercise. Show that a space X has the maximal compact topology if and only if every compact set in X is closed.
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8.1.41 Exercise. Show that every space X with the maximal compact topology is T1 . (Compare this claim with 8.1.38.) The next result presents the behavior of compactness with respect to the topological product. 8.1.42 Theorem. (Tychonoff) Let {Xλ }λ∈Λ be a nonempty family of nonempty topological spaces and consider the topological product X = Πλ∈Λ Xλ . Then X is compact if and only if Xλ is for each λ ∈ Λ. Proof: Since the projection pλ : X −→ Xλ is continuous and surjective, then, if X is compact, so is Xλ is compact for each λ ∈ Λ. Conversely, let Xλ be compact for each λ ∈ Λ. Let U be an ultrafilter in X. By Proposition 7.5.5, the image of U under the projection, pλ (U), is an ultrafilter in Xλ and since Xλ is compact, pλ (U) converges. Consequently, by 7.5.1, U converges and therefore X is compact. t u By the Tychonoff Theorem 7.5.4 we have the following result. 8.1.43 Corollary. Let {Xλ }λ∈Λ be a nonempty family of nonempty topological sets and take their topological product X = Πλ∈Λ Xλ . Then X is a compact Hausdorff space if and only if Xλ is a compact Hausdorff space for each λ ∈ Λ. t u Exercise 8.1.8 provides an example of an infinite product of discrete spaces which is not discrete, although it is compact. Since this product has an infinite underlying Compact.sets in a topological spaceset, it is impossible that it is discrete, because it would not be compact. Compact sets in a topological space have in many aspects a similar behavior to points. The next results show this behavior. In finite products we have the following. 8.1.44 Theorem. Let X and Y be topological spaces. If A ⊂ X and B ⊂ Y are compact and W is a neighborhood of A×B in X ×Y , Then there are neighborhoods U of A in X and V of B in Y such that U × V ⊂ W . b of x in X Proof: For every point (x, y) ∈ A × B there are open neighborhoods U b × Vb ⊂ W . For a fixed x ∈ A, since B is compact, and Vb of y in Y such that U b1 , U b2 , . . . , U bk de x, as well as there are finitely many of these neighborhoods, say U S corresponding neighborhoods Vb1 , Vb2 , . . . , Vbk such that B ⊂ Vb 0 = ki=1 Vbi .
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b 0 = Tk U b b0 b0 If we write U i=1 i , then U is an open neighborhood of x and V is an b 0 × Vb 0 ⊂ W . Since A is compact, there are open neighborhood of B such that U b0, U b0, . . . , U b 0 as well as corresponding finitely many of these other neighborhoods U 1 2 l b 0 × Vb 0 ⊂ W and Vb10 , Vb20 , . . . , Vbl0 such that Vbj0 is an open neighborhood of B, U j j Sl b 0 Tl b 0 A ⊂ U = j=1 Uj . Then U and V = j=1 Vj re open neighborhoods of A and B, respectively, such that U × V ⊂ W . t u 8.1.45 Theorem. Let X be a Hausdorff space. If A, B ⊂ X are disjoint compact sets, then there exist disjoint open sets U, V ⊂ X such that A ⊂ U and B ⊂ V . Proof: Let ∆ ⊂ X × X be the diagonal. Since X is Hausdorff, by 7.1.30, ∆ is closed. Since A and B are disjoint, then A × B ⊂ W = X × X − ∆, and W is an open neighborhood of A × B. Hence, by 8.1.44, there are open sets U, V ⊂ X such that A × B ⊂ U × V ⊂ W . Hence A ⊂ U , B ⊂ V and U and V are disjoint (since their product does not meet the diagonal). t u The following is an immediate consequence. 8.1.46 Corollary. Let X be a compact Hausdorff space. If A, B ⊂ X are disjoint closed sets, then there exist disjoint open sets U, V ⊂ X such that A ⊂ U and B ⊂V. t u The separability property for closed sets stated in the previous corollary is known as normality. In other words, every compact Hausdorff space is normal. This property will be analyzed in detail in the next chapter (9.1). 8.1.47 Exercise. Let X be a Hausdorff space. Show that a finite union of compact subsets of X is compact. Next exercise shows the relationship between compactness and topological sums. 8.1.48 Exercise. Show that a topological sum X = if and only if Λ is finite and each Xλ is compact.
`
λ∈Λ Xλ
is a compact space
8.1.49 Exercise. Let X be a compact Hausdorff space. Show that for every T point x ∈ X the connected component Cx , that contains x, coincides with {C ⊂ X  C es abierto y cerrado y C ⊃ Cx }. (Hint: Use 8.1.45.)
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8.1.50 Exercise. Let X be a connected, compact and Hausdorff space. Prove that for every set A ⊂ X there exists a compact connected set C containing A, which is minimal, i.e. such that if D is a compact connected set and A ⊂ D ⊂ C, then D = C. (Hint: Use Zorn’s lemma.) 8.1.51 Exercise. Show that every locally connected compact set has finitely many connected components. (Hint: Use Exercise 6.2.3.) What happens if X is not locally connected? (Hint: Analyze the Cantor set. See 2.2.7.) 8.1.52 Exercise. Let X be a compact metric space and let f : X −→ X be an isometry, i.e. a map such that d(f (x), f (y)) = d(x, y). Prove that f is surjective. (Hint: If one takes y ∈ X, then the sequence y, f (y), f (f (y)), . . . has points as close to y as desired.) To finish this section, we shall give an interesting characterization of compact spaces. For that we require the following. 8.1.53 Definition. Let f : X −→ Y be a continuous map. Consider the following property of f : (F) For every filter F en X and for every cluster point y ∈ Y of the image filter f (F) there exists a cluster point x ∈ X of F such that f (x) = y. In this case we simply say that f has property (F). 8.1.54 Lemma. Let f : X −→ Y be a continuous map. If f has property (F), then f is closed. Proof: Let A ⊂ X be a nonempty closed set and y ∈ f (A). If FA denotes the filter of all supersets of A in X, then its image f (FA ) is the filter of the supersets of f (A) in Y . Since y ∈ f (A) ⊂ G for any G ∈ f (FA ), y is a cluster point of f (FA ). By property(F), there exists a cluster point x ∈ X of FA such that f (x) = y. t u. Hence x ∈ A = A and y = f (x) ∈ f (A), which shows that f (A) is closed. 8.1.55 Lemma. Let fλ : Xλ −→ Yλ , λ ∈ Λ, be a family of maps with property Q Q Q (F), then the product map f = fλ : X = Xλ −→ Y = Yλ also has property (F). Therefore it is a closed map. Proof: Assume that each fλ has property (F). Let F be a filter in X and y = (yλ ) ∈ Y a cluster point of f (F). Since the projection qλ : Y −→ Yλ is continuous, then
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yλ is a cluster point of qλ f (F) = fλ pλ (F), where pλ : X −→ Xλ is the projection. Since fλ has property (F), the image pλ (F) has a cluster point xλ ∈ Xλ such that fλ (xλ ) = yλ . By 7.5.6(c), x = (xλ ) ∈ X, which is such that f (x) = y, is a cluster point of F. Thus f has property (F). t u 8.1.56 Theorem. Let X be a topological space. The X is compact if and only if for every space Z the projection projZ : X × Z −→ Z is a closed map. Proof: Assume first that X is compact. If P denotes the singular space, then clearly the only map X −→ P has property (F). It is also clear that idZ : Z −→ Z has property (F), so that by 8.1.55, also the product of both maps, namely, the projection projZ : X × Z −→ Z has property (F). Thus it is a closed map. Conversely assume that projZ : X × Z −→ Z is a closed map for any space Z. Let F be a filter in X and let XF∗ be as in 7.1.44 (constructed for the set X). Take ∆ = {(x, x)  x ∈ X} ⊂ X × XF∗ and put F = ∆ ⊂ X × XF∗ . Since by assumption projXF∗ : X × XF∗ −→ XF∗ is closed, then projXF∗ (F ) is closed in XF∗ . Since X ⊂ projXF∗ (F ) and by 7.1.45, X is not closed in XF∗ , then ∞ ∈ projXF∗ (F ), i.e. thete exists x ∈ X such that (x, ∞) ∈ F . Hence every neighborhood V of x in X and for each element M ∈ F, the neighborhood V × (M ∪ {∞} of (x, ∞) in X × XF∗ is such that V × (M ∪ {∞}) ∩ ∆ = (V × M ) ∩ ∆ 6= ∅. In other words V ∩M = 6 ∅, so that x is a cluster point of F. Therefore, by 8.1.4(C000 ), X is compact. t u 8.1.57 Exercise. Let X a noncompact Hausdorff space and define A = {A ⊂ X  X − A is compact} ∪ {∅} . Show that A is a topology in X, Called cocompact topology. How does this topology compare with the original topology of X?
8.2
Compactness and countability
Threr are interesting relationships between compactness and several countability properties. In this section we shall study the property of being compact with respect to the behavior of sequences. We shall prove the Lindel¨of theorem on the existence of countable subcovers of a given open cover and we shall use the proven results to finally prove the theorems of Heine–Borel–Lebesgue and Bolzano– Weierstrass.
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8.2.1 Definition. A topological space X is said to be countably compact if every sequence in X has a cluster point. The space X is said to be sequentially compact if every sequence in X has a convergent subsequence. We have the following. 8.2.2 Proposition. (a) Every compact space X is countably compact. (b) Every sequentially compact space X is countably compact. Proof: (a) This is clear. Namely, since X is a compact space, every filter in X has a cluster point. In particular, every elementary filter in X has a cluster point, i.e. every sequence in X has a cluster point. (b) This is obvious.
t u
The converse of (b) is not necessarily true. However one has the following. 8.2.3 Theorem. Every countably compact firstcountable space X is sequentially compact. Proof: Take a sequence {xn } in X and let x be a cluster point of the sequence. Since X is firstcountable, we may take a countable neighborhood basis of x, U1 ⊃ U2 ⊃ U3 ⊃ · · · ⊃ Un ⊃ · · · . Since x is a cluster point of {xn }, then {xn  n > m} ∩ Uk 6= ∅ for all k and m. Choose xn1 ∈ {xn } ∩ U1 . Inductively, assume that we we have chosen xni ∈ Ui , n1 < n2 < · · · < nk and then choose xnk+1 ∈ {xn  n > nk } ∩ Uk+1 . Hence the sequence {xnk } is a subsequence of {xn } and clearly xnk → x. t u 8.2.4 Theorem. (Lindel¨of) Assume X to be a secondcountable space. Then X satisfies the Lindel¨of axiom (L) Every open cover of X contains a countable subcover. Proof: Let {Qn } be a countable basis for the topology of X and let {Uλ } be an open cover of X. For each λ one has [ Uλ = Qn . Qn ⊂Uλ
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Let n1 < n2 < · · · be such that Qni ⊂ Uλ for some λ. Thus ∞ [
Qni =
i=1
[
Uλ = X .
λ∈Λ
Now take Uλi such that Qni ⊂ Uλi . Clearly ∞ [
Uλi = X .
i=1
Therefore {Uλi } is a countable subcover of X.
t u
of 8.2.5 Definition. A topological space X that satisfies (L) is called a Lindel¨ space. 8.2.6 Lemma. A space X is countably compact if and only if every countable open cover of X contains a finite subcover. Proof: Let X be countably compact and let {Un } be a countable open cover of X. Assume on the contrary that {Un } does not contain a finite subcover. Thus S Vn = nk=1 Uk 6= X for all n. Take xn ∈ X − Vn . Since X is countably compact, {xn } has a cluster point x ∈ X. Then x ∈ Um for some m. On the other hand Um ⊂ Vm . Therefore Vm is a neighborhood of x that contains only finitely many terms of the sequence. This is a contradiction to the fact that x is a cluster point of the sequence. Hence {Un } contains a finite subcover. Conversely, assume that every countable open cover of X contains a finite subcover. Assume that there is a sequence {xn } in X with no cluster point. Hence any point x ∈ X has a neighborhood V such that V ∩ {xn  n ≥ k} = ∅ for k sufficiently large. Take the open set Vk = X − {xn  n ≥ k}. Clearly V1 ⊂ V2 ⊂ · · · ⊂ Vk ⊂ · · · . Then {Vk } es una countable open cover of X, since for any x ∈ X there is a neighborhood V of x such that V ∩ {xn  n ≥ k} = ∅ for some k Hence x 6∈ {xn  n ≥ k}. Thus x ∈ Vk . The countable open cover {Vk } does not contain a finite subcover since Vk 6= X for all k. t u 8.2.7 Note. Many authors define a countably compact space X as a space such that each countable open cover contains a finite subcover. The previous lemma shows that our definition is equivalent. 8.2.8 Proposition. If X is a countably compact Lindel¨ of space, then X is compact.
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Proof: Let {Uλ } be an open cover of X. By the Lindel¨of axiom, {Uλ } contains a countable subcover, and by Lemma 8.2.6, this countable cover contains a finite subcover. t u As a consequence of the Lindel¨of Theorem and of 8.2.8 we obtain the following. 8.2.9 Theorem. Let X be a secondcountable, countably compact space. Then X is compact. t u 8.2.10 Theorem. Let X be a countably compact metric space. Then X satisfies the following: (a) For every open cover {Qλ } of X there exists δ > 0, such that for any point x ∈ X there is some λ such that the Dδ (x) ⊂ Qλ . (Such a δ is called Lebesgue number of the cover.) (b) For every ε > 0 there exist points x1 , . . . , xk ∈ X such that X = Dε (x1 ) ∪ · · · ∪ Dε (xk ). (c) X is compact. (d) X is sequentially compact. Proof: (a) Let {Qλ } be an open cover which does not admit a Lebesgue number and let δn = 1/2n . Hence there exists a point xn such that Dδn (xn ) 6⊂ Qλ for all λ. Since X is countably compact, the sequence {xn } has a cluster point x ∈ X and there is a κ ∈ Λ such that x ∈ Qκ . But Qκ is open and so there is m ∈ N such that Dδm (x) ⊂ Qκ . For any point y ∈ Dδm+1 (x) one has d(z, y) ≤
1 1 1 1 =⇒ d(z, x) ≤ m+1 + m+1 = m . 2m+1 2 2 2
This means Dδm+1 (y) ⊂ Dδm (x) ⊂ Qκ and therefore, for all n ≥ m + 1, xn 6∈ Dδm+1 (x). This contradicts the fact that x is a cluster point. (b) If this were false, then there woud be an ε > 0such that X is not covered by finitely many balls of radius ε. Take any point x1 ∈ X. The ball Dε (x1 ) does not cover X. Now take a point x2 ∈ X − Dε (x1 ). Assume inductively that we have taken points x1 , . . . , xk such that xi 6∈ Dε (x1 ) ∪ · · · ∪ Dε (xi−1 ), i = 2, . . . , k. S S The union ki=1 Dε (xi ) does not cover X. Hence take xk+1 ∈ X − ki=1 Dε (xi ). The sequence {xn } cluster points since given any point x ∈ X, the ball Dε/2 (x) contains at most one point of the sequence, because, by construction, the distance between two of them is greater than or equal to ε.
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(c) Let {Qλ } be an open cover of X. By (a), this cover has a Lebesgue number δ. By (b), we know that X is covered by finitely many balls of radius δ, namely S X = ki=1 Dδ (xi ). Since δ is a Lebesgue number of {Qλ }, for each xi there is a λi S such that Dδ (xi ) ⊂ Qλi . Therefore X = ki=1 Qλi . (d) This is a consequence of 8.2.3, since every metric space is firstcountable.u t The following is a basic result. 8.2.11 Theorem. The unit interval I = [0, 1] is compact. Proof: By the previous theorem, it is enough to prove that I is countably compact, since it is metric. So take a sequence {xn } in I. For each point c ∈ I put Sc = {n  xn < c} and consider the set C = {c  Sc is finite}. Clearly 0 ∈ C. Therfore there exists c0 = sup C. We shall see that c0 is a cluster point of {xn }. Take ε > 0. It is obvious that Sc0 −ε/2 is finite, so that xn > c0 − ε for infinitely many values of n. If c0 = 1, then the ball with center c0 and radius ε in I contains a tail of the sequence. Assume now that c0 < 1. Since c0 = sup C, c0 + ε 6∈ C. Hence Sc0 +ε is infinite. Consequently, also in this case the ball with center c0 and radius ε in I contains points xn for infinitely many values of n. Thus c0 is a cluster point of {xn } t u As a consequence of 8.1.30, the Tychonoff Theorem and the previous theorem, we can finally prove the two statements at the beginning of the chapter (8.1.1 and 8.1.2). 8.2.12 Theorem. (Heine–Borel–Lebesgue) A subset A ⊂ Rn is compact if and only if it is closed and bounded. Proof: If A is compact, then by 8.1.30 it is closed and bounded, since Rn is metric. Conversely assume that A is closed and bounded. Since A is bounded, there is an interval J = [a, b] ⊂ R such that A ⊂ J n ⊂ Rn . Moreover, there is a homeomorphism J ≈ I, so that J is compact too. By the Tychonoff Theorem 8.1.42, J n is compact, and since A is closed in Rn , it is also closed in J n . But since J n is compact, A is compact. t u As a consequence of the Heine–Borel–Lebesgue Theorem we obtain the other statement at the beginning of the chapter.
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8.2.13 Theorem. (Bolzano–Weierstrass) A subset A ⊂ Rn is closed and bounded if and only if every sequence in A has a cluster point in A. Proof: By 8.2.12, A is closed and bounded if and only if A is compact, and by 8.2.10, since A is metric, A is compact if and only if A is countably compact. t u 8.2.14 Exercise. (a) Similarly to the proof of 8.2.11, show that any totally ordered set with a maximal and a minimal element, and satisfying the supremum axiom, becomes a countably compact space if it is furnished with the order topology. (b) Prove that it is, in fact, compact. (Hint: Use open cover.) There is a local version of the compactness concept, which is quite useful, in case that the considered space is not compact. 8.2.15 Definition. A topological space X is locally compact if for every point x ∈ X there is a compact neighborhood U ∈ Nx . Many authors define a locally compact space adding the condition that it is also a Hausdorff space. This assumption facilitates the statements of several results. We shall add this condition explicitly each time that we need it. The following is clear. 8.2.16 Proposition. Every compact space X is locally compact.
t u
8.2.17 Examples. (a) A topological nmanifold is a Hausdorff secondcountable space X such that each of its points has a neighborhood U which is homeomorphic to an open set in Rn . Then every topological nmanifold is a a locally compact space. Q ω (b) Take Rω = ∞ i=1 Ri , where Ri = R and take x = (xi ) ∈ R . Then any Q neighborhood of x contains a neighborhood of the form Qi , where Qi is neighborhood of xi and Qi = R for almost every index i. By the Tychonoff Theorem, a neighborhood such as this cannot be compact. Hence no other neighborhood of x would be compact and thus Rω is not locally compact.
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8.2.18 Definition. A topological space X is regular if for any point x ∈ X, the closed neighborhoods of x constitute a neighborhood basis. In other words, X is regular if and only if for every x ∈ X and every V ∈ Nx , there W ∈ Nx such that W is closed and W ⊂ V . In the next chapter we shall study this property in more detail (9.1.2). 8.2.19 Theorem. Every locally compact Hausdorff space X is regular. Proof: Since every compact set in a Hausdorff space X is closed, it is enough to prove that the compact neighborhoods of a point x ∈ X build up a neighborhood basis of x, if x is locally compact and Hausdorff. Since X is Hausdorff, the set T {x} = {V ∈ Nx  V is closed}. The closed neighborhoods of x constitute a filter basis of a filter F which is coarser than the neighborhood filter Nx of x. But, on the other hand, F is finer than Nx , since by definition of a cluster point of a filter, x is the only cluster point of F. Since X is compact, by 8.1.6, F converges to x. This proves that the closed (compact) neighborhoods of x en X build up a neighborhood basis of x. If X is not compact, but only locally compact, then each x ∈ X has a compact neighborhood V . Thus V is closed and so x has a neighborhood basis relative to V whose elements are compact neighborhoods. Since V itself is a neighborhood, then this neighborhood basis in V is also a neighborhood basis in X. t u 8.2.20 Corollary. A Hausdorff space X is locally compact if and only if each point of X has a basis of compact neighborhoods. t u Some authors define a locally compact space X as space such that each of its points has a basis of compact neighborhoods. The latter corollary shows that for Hausdorff spaces their and our definition are equivalent. The property of a space being locally compact is not inherited by subspaces. 8.2.21 Example. R is locally compact but Q ⊂ R is not locally compact. We have though the following statements. 8.2.22 Proposition. Let X be a locally compact space A ⊂ X a closed subset. Then A is locally compact. Proof: Take x ∈ A and let V be a compact neighborhood of x in X. Then V ∩ A is a compact neighborhood x in A. t u
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8.2.23 Proposition. Let X be a locally compact Hausdorff space and A ⊂ X an open subset. Then A is locally compact. Proof: Since the compact neighborhoods of a point x ∈ X build up a neighborhood basis in X, those neighborhoods which are contained in A are a neighborhood basis of x in A. t u 8.2.24 Definition. Let X be a topological space and take A ⊂ X. We say that A is locally closed in X if A = Q ∩ C, where Q is open and C is closed in X. 8.2.25 Exercise. Show that A is locally closed in X if and only if for every x ∈ A there is a neighborhood V of x in X such that A ∩ V is closed in V . We can put together the statements of Propositions 8.2.22 and 8.2.23, in the case of Hausdorff spaces. Namely, we have the following result. 8.2.26 Theorem. Let X a locally compact Hausdorff space and let A ⊂ X be locally closed. Then A is locally compact. Proof: By definition, A = Q ∩ C with Q open and C closed in X. By 8.2.22, C is locally compact and, since A is open in C because it is the intersection of C with an open set Q, by 8.2.23, A is locally compact. t u 8.2.27 Theorem. Let {Xλ } be a nonempty family of nonempty topological spaces. Q The topological product X = Xλ is locally compact if and only if all factors are locally compact and all but a finite number of them are compact. Proof: Assume first that X is locally compact. Therefore any point x = (xλ ) has a compact neighborhood V that contains another neighborhood Q of the form Q Q = Qλ , with Qλ neighborhood of xλ and Qλ = Xλ for all but a finite number of indexes λ. If pλ : X −→ Xλ is the projection, then the image pλ (V ) is a compact neighborhood of xλ , since Qλ ⊂ pλ (V ). Hence xλ , which can be arbitrarily chosen, has a compact neighborhood, i.e. Xλ is locally compact. Moreover, pλ (V ) = Xλ all but a finite number of indexes λ, hence, for all these indexes, Xλ is compact. Conversely, assume that Xλ is locally compact for all λ and that Xλ is compact for all but a finite number of indexes λ. Let us say that λ = κ1 , . . . , κn are these indexes. Take x = (xλ ) ∈ X and take a compact neighborhood Vκi of xκi . Take Q Vλ = Xλ for λ 6= κi , i = 1, . . . , n. Then it is clear that V = Vλ is a compact neighborhood of x in X. t u
8.3 The Alexandroff compactification
8.3
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The Alexandroff compactification
In this section we shall study the simplest of all compactifications of a space, which is obtained by adding to the space only one point. We start with an example. 8.3.1 Example. We can embed Rn into the nsphere Sn via the map i : Rn −→ Sn given by i(x1 , . . . , xn ) =
(2x1 , 2x2 , . . . , 2xn , x21 + · · · + x2n − 1) . x21 + · · · + x2n + 1
This map is the inverse of the stereographic projection, and it embeds Rn as the complement of the north pole (0, . . . , 0, 1) of Sn . Namely, Rn becomes a subspace of Sn such that its complement is one point (see 2.6.6).
π
Figure 8.1 The stereographic projection In the example above we observe that Rn , which is a noncompact space, can be completed to a compact space by just adding one extra point. Indeed we obtain Sn after adding to Rn a point at infinity. We can easily prove that Rn seen as a subspace of Sn is dense. 8.3.2 Definition. Let X be a noncompact topological space. A compactification of X is a space X 0 together with an embedding i : X −→ X 0 such that the image i(X) is dense in X 0 . Hence Example 8.3.1 shows that the inverse of the stereographic projection i : Rn −→ Sn is a compactification of Rn . 8.3.3 Definition. Let X be a Hausdorff space and consider the set X ∗ consisting of the union of X with an additional point ∞. Namely X ∗ = X ∪ {∞}. Take the topology on X ∗ given by A∗ = A ∪ {A ⊂ X ∗  ∞ ∈ A, X ∗ − A ⊂ X is compact} , where A is the original topology on X. The resulting topological space X ∗ is called the Alexandroff construction of X.
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8.3.4 Example. If X is discrete and infinite and F is the cofinite filter, built up with the cofinite sets in X, 7.1.10(f), then the Alexandroff construction of X ∗ coincides with the space XF∗ associated to the filter F after 7.1.44 (see 8.3.8). Next we see that A∗ is indeed a topology on X ∗ . 8.3.5 Proposition. (a) A∗ is a topology on X ∗ that induces on X ⊂ X ∗ the original topology A. (b) X ∗ with the topology A∗ is a compact space. (c) X is open in X ∗ . Hence, if X is a noncompact space, then its Alexandroff construction X ∗ , together with the natural inclusion is a compactification. Proof: (a) We have to show that arbitrary unions and finite intersections of elements of A∗ that contain ∞ are again elements of A∗ . Let {Aλ } be a family of elements of A∗ such that ∞ ∈ Aλ for all λ. Take A = ∪Aλ . Then ∞ ∈ A and X ∗ − A = ∩(X ∗ − Aλ ), which is compact since each X ∗ − Aλ is compact too and X is S Hausdorff. Then Aλ ∈ A∗ . Thus A∗ is a topology on X ∗ . Take now elements A1 and A2 of A∗ such that ∞ ∈ A1 ∩ A2 . Then X ∗ − (A1 ∩ A2 ) = (X ∗ − A1 ) ∪ (X ∗ − A2 ), which is clearly compact. Therefore A1 ∩ A2 ∈ A∗ . (b) Let {Qλ } be an open cover of X ∗ . Take λ0 such that ∞ ∈ Qλ0 . The set C = X ∗ − Qλ0 ⊂ X is compact and {Qλ } is an open cover of C in X ∗ . Hence there is a finite subcover of C, say consisting of Qλ1 , . . . , Qλk . One clearly has that the finite family {Qλ0 , Qλ1 , . . . , Qλk } is a cover X ∗ . Thus X ∗ is compact. (c) This is clear, since X ∈ A. If X is noncompact, then the closure X = X ∗ , since X and X ∗ differ by one point and X must contain points not in X. In other words, the closure X contains {∞}. t u 8.3.6 Definition. Let X be a noncompact space. The space X ∗ is called the Alexandroff compactification of X. In some texts it is also called onepoint compactification of X. The assumption that X is Hausdorff is not essential. In the same way that we proved 8.3.5, one can prove the next result.
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8.3 The Alexandroff compactification
8.3.7 Theorem. Let X be a topological space with topology A and let X ∗ = X ∪ {∞}. Define A∗ = A ∪ {A ⊂ X ∗  ∞ ∈ A,
X ∗ − A is compact and closed} .
Then (a) A∗ is a topology in X ∗ that has X as a subspace (b) X ∗ with the topology A∗ is compact, and (c) if X is not compact, then it is dense in X ∗ . 8.3.8 Exercise. Compare the given construction of X ∗ with the construction of XF∗ given in 7.1.44, and show that if X is discrete, both resulting spaces coincide. 8.3.9 Exercise. Let X be a compact space. Show that the Alexandroff construction X ∗ of X, 8.3.3, yields the topological sum of X and a point, i.e. X ∗ = X t {∞} . Hence when X is compact, X is not dense in X ∗ . Even when the space X is Hausdorff, the Alexandroff compactification X ∗ of X is not necessarily Hausdorff. 8.3.10 Example. Let Q ⊂ R have the relative topology. Then Q is a Hausdorff space. However, Q∗ is not a Hausdorff space. Namely, it is not possible to separate with open neighborhoods ∞ from any other element q ∈ Q. To see this, take q ∈ Q. A typical neighborhood of q is an interval V = (a, b) ∩ Q. Let A be a neighborhood of ∞ in Q∗ . Therefore C = Q∗ − A is compact. If V ∩ A = ∅, then V ⊂ C. Notwithstanding, this is impossible since there are sequences in V (o rational numbers) which in R converge to irrational numbers. Hence they do not converge in Q, thus contradicting the compactness of C. 8.3.11 Exercise. Let Q0 be the quotient R/R − Q and take the composite e:Q
ÂÄ
/R
/ / R/R − Q .
(a) Show that e : Q −→ Q0 is an embedding. (b) Show that Q is dense in Q0 . (c) Is Q0 compact?
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8 Compactness
(d) How do Q∗ and Q0 compare, if Q∗ is as in Example 8.3.10? There is a way to guarantee that the Alexandroff compactification of a Hausdorff space is again a Hausdorff space. We have the following result. 8.3.12 Theorem. (Alexandroff) X is a Hausdorff locally compact space if and only if its Alexandroff compactification X ∗ is a Hausdorff compact space. Proof: First we prove that if X is Hausdorff locally compact, then X ∗ is Hausdorff. To see this it is enough to prove that given x ∈ X there are a neighborhood of x and a neighborhood of ∞ which are disjoint. Since X is locally compact, there is a compact neighborhood V of x ∈ X. Take W = X ∗ −V . Then W is a neighborhood of ∞ in X ∗ which is obviously disjoint to V . Conversely, if X ∗ is Hausdorff, then X is also Hausdorff, because it is a subspace. Moreover, if we take x ∈ X, then there are disjoint neighborhoods V of x and W of ∞ in X ∗ . We may assume that W is open, so that X ∗ − W is closed and compact in X. Since V ⊂ X ∗ − W , then V ⊂ X ∗ − W and thus V is a compact neighborhood of x in X, and so X is locally compact. t u 8.3.13 Remark. Since X is open in X ∗ , if X ∗ is firstcountable, then X is also firstcountable. Moreover, ∞ has a countable neighborhood basis. On the other hand, if X ∗ is secondcountable, then the open sets of a countable basis of the topology of X ∗ that contain ∞, form a countable neighborhood basis of ∞. This brings us to the next definition. 8.3.14 Definition. A toplogical space X is countable at infinity if ∞ ∈ X ∗ has a countable neighborhood basis. 8.3.15 Proposition. The Alexandroff construction X ∗ is first, resp. secondcountable, if and only if X is first, resp. secondcountable and it is countable at infinity. t u Now we shall study the meaning of the countability at infinity. Let X Hausdorff locally compact space and let V be a openneighborhood basis of ∞ in X ∗ . The complements of the neighborhoods in V are compact sets in X and since X ∗ is T T Hausdorff, then {∞} = V ∈V V . Hence X = X ∗ − {∞} = X ∗ − V ∈V V = S ∗ V ∈V (X − V ), i.e. X is a union of compact sets, one for each element of V. Hence if V is countable, i.e. if X is countable at infinity, then we have that X is a countable union of compact sets.
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8.3 The Alexandroff compactification
S Conversely, assume that X = ∞ n=1 Kn , where Kn is a compact set for each n. ∗ Therefore {X − Kn } is a countable family of open neighborhoods of ∞ in X ∗ . We shall use these neighborhoods to construct a countable neighborhood basis of ∞. Since by 8.2.19 every Hausdorff (locally) compact space is regular, there is a compact neighborhood V1 of ∞ in X ∗ , such that V1 ⊂ X ∗ − K1 . Now we have that (X ∗ − K2 ) ∩ V1◦ is an open neighborhood of ∞ in X ∗ too. For the same reason as before, this neighborhood contains a compact neighborhood V2 of ∞. Inductively assume that we have already constructed a neighborhood Vn of ∞ such ◦ . Using the same argument as before, there is an open that Vn ⊂ (X ∗ − Kn ) ∩ Vn−1 neighborhood Vn+1 of ∞ such that Vn+1 ⊂ (X ∗ − Kn+1 ) ∩ Vn◦ . Thus Vn+1 ⊂ Vn◦ for all n. We now see that {Vn } is a neighborhood basis of ∞. For this, let Q be an open neighborhood of ∞. Hence X ∗ − Q is compact. On the other hand we have ∞ [
∗
∗
(X − Vn ) = X −
n=1
⊃ X∗ −
∞ \
Vn
n=1 ∞ \
(X ∗ − Kn )
n=1 ∗
∞ [
∗
= X − (X −
Kn )
n=1
=
∞ [
Kn = X .
n=1
Hence {X ∗ − Vn } is an open cover of the compact set X ∗ − Q. Besides, since X ∗ − V1 ⊂ X ∗ − V2 ⊂ · · · , we have X ∗ − Q ⊂ X ∗ − Vn for n large enough, i.e. Vn ⊂ Q for some n. This proves that {Vn } is a neighborhood basis of ∞. S In fact, we have proved that X = ∞ n=1 Qn . Thus we have the following. 8.3.16 Theorem. Let X be a locally compact Hausdorff space, then the following ar equivalent: (a) X is countable at infinity. (b) X is a countable union of compacto sets. S (c) X = ∞ n=1 Qn , where Qn is open for all n, Qn is compact, and Qn ⊂ Qn+1 , n ∈ N. t u 8.3.17 Note. Notice that the fact that X is a countable union of compact sets does not imply that X is locally compact. For example, consider X = Q.
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8 Compactness
Now we analyze an important example, where several of the concepts introduced in this chapter play a role. 8.3.18 Example. The Alexandroff compactification C∗ of the complex plane C, which by 8.3.1 is homeomorphic to the 2sphere, is the socalled Riemann sphere. If we now take the 3sphere S3 ⊂ C2 ≈ R4 , defined by
S3 = {(z, w) ∈ C2  kzk2 + kwk2 = 1} , there is a continuous surjective map 3
∗
p : S −→ C ,
½z p(z, w) =
if w 6= 0 w ∞ if w = 0 .
Since S3 is compact and C∗ is Hausdorff, then p is an identification. Define the complex projective space CPn as the quotient space S2n+1 /∼, where S2n+1 ⊂ Cn+1 ≈ R2n+2 . Hence, if x = (x1 , . . . , xn+1 ), y = (y1 , . . . , yn+1 ) ∈ S2n+1 , x = λy for λ ∈ C, we can take the identification q : S2n+1 −→ CPn (compare with 4.2.17(d)). Then, in particular, the map p : S3 −→ C∗ defined above is compatible with the identification q : S3 −→ CP1 and hence we obtain a homeomorphism ϕ : CP1 −→ C∗ ≈ S2 . When we take CPn as a quotient of S2n+1 , what we in fact are doing is taking an arbitrary point y ∈ S2n+1 and the orbit {λy  λ ∈ S1 } ⊂ S2n+1 , and identifying each of these orbits in one point. (In other words, we have an action of the group S1 on the (2n + 1)sphere and CPn is the orbit space of this action, as we already explained in 5.5.15.5.5. The only difference is that there we considered the action of finite groups and here we are dealing with the action of the group S1 which has a topological structure involved in the action.) Concluding we may say that there is a map (identification) η : S3 −→ S2 , such that its fiber, i.e. the inverse image of each point, is a copy of S1 , which plays an important role in homotopy theory and is known as the Hopf fibration. 8.3.19 Exercise. In the previous example, prove that all assertions that we made hold. 8.3.20 Exercise. Prove that the map ϕ : S3 −→ S2 ⊂ C × R = R3 such that ϕ(z, w) = (2zw, kzk2 − kwk2 ) coincides with the Hopf fibration η defined above. 8.3.21 Exercise. Prove that there is a canonical map γ : RP2n+1 −→ CPn , so that the diagram
S2n+1II
t p ttt t t t yty tt
RP2n+1
γ
II q II II I$ $ / CPn ,
189
8.3 The Alexandroff compactification
commutes, where p and q are the canonical identifications. Show that ϕ is surjective and that its fiber γ −1 (x) ≈ S1 . More precisely, notice that there is an action of S1 on RP2n+1 , i.e. a way of multiplying the elements of RP2n+1 by unit complex numbers, in such a way that the corresponding orbit space is CPn . (Hint: For the latter statement, consider the action ζ 2 · [z] = [ζz], where ζ ∈ S1 , z ∈ S2n+1 ⊂ Cn+1 ≈ R2n+2 and [z] = p(z) ∈ RP2n+1 . Notice that ζ 2 is any element of S1 that determines ζ up to sign.) 8.3.22 Exercise. Prove that the map α : S1 × RP2n+1 −→ RP2n+1 given by α(ζ 2 , [z]) = [ζz] is well defined and is such that for any given point z ∈ RP2n+1 , the map S1 −→ RP2n+1 given by ζ 2 7→ [ζz] is an embedding. Moreover prove that if we identify two points [z] and [z 0 ] in RP2n+1 if and only if [z 0 ] = [ζz], then the quotient map is (homeomorphic to) CPn . To close this section, we analyze an interesting example. 8.3.23 Example. As we have already mentioned in 5.4.18, the projective plane RP2 can be obtained from the disk B2 if one identifies antipodal points on the boundary, namely
RP2 = B2 /∼ where x ∼ y ⇔ x, y ∈ S1 = ∂ B2 and x = ±y . If N = (0, 1), S = (0, −1) ∈ S1 ⊂ B2 are the poles, then we have a homeomorphism ϕ : B2 − {N, S} −→ [−1, 1] × (−1, 1) , given by
x1
ϕ(x1 , x2 ) = ( p
1 − x22
, x2 ) ,
with inverse ψ : [−1, 1] × (−1, 1) −→ B2 − {N, S} , given by
p ψ(s, t) = (s 1 − t2 , t) .
If q : B2 −→ RP2 denotes the identification described above and P = q(N ) = q(S), then the homeomorphism ϕ determines a homeomorphism in in the quotient spaces ϕ : RP2 − {P } = q(B2 − {N, S}) −→ [−1, 1] × (−1, 1)/∼ , where (1, t) ∼ (−1, −t). The latter space M◦ = [−1, 1] × (−1, 1)/∼ is the open Moebius band. Hence we have proved that if we take out a point (P ) from the projective plane RP2 we obtain, up to homeomorphism, the open Moebius band M◦ . Consequently, since the projective plane is compact (because it is the image of a compact space) and since RP2 − {P } is dense it, then RP2 is its Alexandroff compactification. We have proved the following result.
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8 Compactness
8.3.24 Proposition. The Alexandroff compactification of the open Moebius band is the projective plane, i.e. (M◦ )∗ ≈ RP2 . t u 8.3.25 Exercise. Using 8.3.23 or 8.3.24, show that the quotient of the (closed) Moebius band M obtained by collapsing the boundary ∂ M onto a pointm is the projective plane. Here we define the boundary by ∂ M = q([−1, 1] × {−1, 1}), if q : [−1, 1] × [−1, 1] −→ M is the identification that defines M, that is, M/∂ M ≈ RP2 .
8.4
Proper maps
In the previous section we defined the Alexandroff compactification of a noncompact topological space. In this section we shall study the behavior of this construction with respect to maps. We shall give several characterizations of the maps that can be extended to the Alexandroff compactifications. We shall call these maps proper. The reader can take a look at other books like [3], where alternative and complementary treatments of this topic are pursued. Here we start analyzing some examples. (In what follows, abusing notation, we denote by ∞ the added point in every space to obtain its compactification, but we shall understand that to different spaces different points ∞ are added.)
8.4.1 Examples. (a) Take X = (0, 1), Y = [0, 1] and the inclusion f : X −→ Y . Clearly the Alexandroff compactification X ∗ ≈ S1 with embedding i : (0, 1) −→ S1 given by i(t) = e2πit . On the other hand, Y ∗ = Y t {∞}. It is an easy exercise to observe that f does not admit an extension f 0 : X ∗ −→ Y ∗ . (b) Moreover, if we consider Z = R and g : X −→ Z, then g = j ◦ f , where j is the inclusion of [0, 1] en R, then we have that both X and Z are not compact. In this case, an extension g ∗ : X ∗ −→ Z ∗ of g does not exist either. (c) Take now X = Y = (0, 1) and let f : X −→ Y be given by f (t) = 21 for all t. In this case, f does admit an extension f 0 : X ∗ −→ Y ∗ , namely the constant map with value 12 or, the map S1 −→ S1 with value −1 ∈ S1 , if we put X ∗ = S1 . However, it is impossible to find an extension f ∗ such that f ∗ (1) = 1, that is, an extension that maps the added point ∞ to ∞.
191
8.4 Proper maps
The previous examples show that not every map between noncompact spaces can be extended canonically to their Alexandroff compactifications, mapping ∞ to ∞. 8.4.2 Exercise. Let f : R −→ R be continuous and let f ∗ : R∗ −→ R∗ be given by f ∗ (t) = f (t) if t ∈ R and f ∗ (∞) = ∞. Show that f ∗ is continuous if and only if for every sequence {tn } in R such that tn → ∞, one has that f (tn ) → ∞. Let X and Y be topological spaces and take a continuous map f : X −→ Y . Define f ∗ : X ∗ −→ Y ∗ by f ∗ (x) = f (x) if x ∈ X and f ∗ (∞) = ∞. f ∗ Is continuous at x ∈ X∗ if and only if for every neighborhood V of f ∗ (x) in Y ∗ , the inverse image (f ∗ )−1 (V ) is a neighborhood of x in X ∗ . Clearly f ∗ is always continuous at x ∈ X (exercise). When is f ∗ continuous at ∞? Let V be an open neighborhood of ∞ in Y ∗ . Hence Y ∗ − V is compact. We wish that (f ∗ )−1 (V ) is an open neighborhood of ∞ in X ∗ . That is, we want that X ∗ − (f ∗ )−1 (V ) is compact. This leads us to the following concept. 8.4.3 Definition. Let X and Y be topological spaces and let f : X −→ Y be a continuous map. We say that f is a proper map if for each compact set K ⊂ Y , the inverse image f −1 (K) ⊂ X is a compact set too. By the previous discussion, the following result must be clear. 8.4.4 Theorem. Let X and Y be topological spaces and let f : X −→ Y be a continuous map. Then f ∗ : X ∗ −→ Y ∗ is a continuous map if and only if f is a proper map. t u Notice that the previous theorem is valid for any spaces X and Y , which can even be compact or not locally compact nor Hausdorff. 8.4.5 Note. Let X be a noncompact space and assume that f : X −→ Y admits an extension f 0 : X ∗ −→ Y . In this case f cannot be proper since f 0 (X ∗ ) ⊂ Y is compact, but X = f 0−1 (f 0 (X ∗ )) is not compact. Let f : X −→ Y be a proper map, i.e. f ∗ : X ∗ −→ Y ∗ is continuous. If A ⊂ X is closed, then A ∪ {∞} is the complement of an open set in X ∗ and hence it is closed also in X ∗ . Thus A ∪ {∞} is compact, so that f ∗ (A ∪ {∞}) = f (A) ∪ {∞}
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8 Compactness
is also compact. If Y is a locally compact Hausdorff space, then Y ∗ is a compact Hausdorff space. Hence f (A) ∪ {∞} is closed in Y ∗ . Since its complement is open, also the complement of f (A) in Y is open. Therefore f (A) is closed in Y . In particular, f (X) is closed in Y and consequently f (X) is locally compact. We have proved the next result. 8.4.6 Proposition. Let Y be a locally compact Hausdorff space and let f : X −→ Y be a proper map. Then f is a closed map and f (X) is locally compact. t u The latter proposition is false if one does not require, at least, that the compact sets of Y ∗ are closed. 8.4.7 Example. Let f be the inclusion of X = {x} in Y = {x, y}, where Y has the indiscrete topology. Obviously f is proper. However, it is not a closed map. The following result gives a useful characterization of proper maps. 8.4.8 Theorem. Let Y be a Hausdorff locally compact space and let f : X −→ Y be continuous. The following are equivalent: (a) f is a proper map. (b) f is a closed map and for every y ∈ Y , f −1 (y) is compact. (c) For every filter F in X and every cluster point y of the image filter f (F), there exists a cluster point x of F such that f (x) = y. Proof: (a)=⇒(b) by 8.4.6 and since every singular space {y} is compact. (b)=⇒(c) Let F be a filter in X and let y ∈ Y be a cluster point of the image filter f (F). Hence y ∈ f (F ) for all F ∈ F. Since f is a closed map, by 4.2.31(d), we have that f (F ) = f (F ). Thus f −1 (y) ∩ F 6= ∅ for every F ∈ F. Therefore the set {f −1 (y) ∩ F  F ∈ F} is a filter basis in f −1 (y) built up by closed sets. Since f −1 (y) is compact, the filter basis has a cluster point x ∈ f −1 (y), i.e. x ∈ F for every F ∈ F. Consequently x is a cluster point of F, as desired. (c)=⇒(a) Let K ⊂ Y be a compact set and let G be a filter in f −1 (K) and F its extension to X. If we denote by fK (G) the image filter f f −1 (K) of G in K, then the image of F under f in Y , f (F), is its extension to Y . Since K is compact, f f −1 (K) has a cluster point y ∈ K, which by 7.4.12(a) is also a cluster point of f (F). By (c), there is a cluster point x ∈ f −1 (y) of F, which, again by 7.4.12(a), is also a cluster point of G. Hence f −1 (y) is compact and therefore f is proper. t u
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8.4 Proper maps
8.4.9 Note. The implication (a)=⇒(b) in the previous theorem is the only one that requires the assumption that Y is Hausdorff and locally compact. The implication (c)=⇒(b), analogously to what was done for (b)=⇒(c), can be shown without that assumption. As a complement to Theorem 8.4.8 solve the next. 8.4.10 Exercise. Let Y be a Hausdorff locally compact space. Show that a map f : X −→ Y is proper if and only if for any ultrafilter U in X and any limit point y ∈ Y of the basis of image ultrafilter f (U), there exists a limit point x ∈ X of U such that f (x) = y. Since Y is Hausdorff, one has in fact that f is proper if and only if for any ultrafilter U in X such that the basis of the image ultrafilter f (U) is convergent, one has that U itself is convergent. There is another interesting characterization of a proper map. 8.4.11 Theorem. Let Y be a Hausdorff locally compact space and let f : X −→ Y be a continuous map. Then f is proper if and only if for any topological space Z the product map f × idZ : X × Z −→ Y × Z is a closed map. Proof: Assume first that f is proper. By 8.4.8(c), f has property 8.1.53(F). Therefore, by 8.1.54, f × idZ : X × Z −→ Y × Z has the same property, thus by 8.1.55, f × idZ is closed. Conversely, let K ⊂ Y be compact. If f × idZ : X × Z −→ Y × Z is a closed map, then the restriction (f × idZ )K×Z = fK × idZ : (f × idZ )−1 (K × Z) = f −1 (K) × Z −→ K × Z is also a closed map (see 4.2.33). Since K is compact, by 8.1.56also the map projZ : K × Z −→ Z is closed. Thus the composite f −1 (K) × Z
fK ×idZ
/K ×Z
projZ
/Z,
which coincides with projZ : f −1 (K) × Z −→ Z, is also a closed map. Again, by 8.1.56, we conclude that f −1 (K) is compact. t u 8.4.12 Exercise. Show that if X is a compact space, then any continuous map f : X −→ Y is proper. Moreover, if P is a singular space, show that g : X −→ P is proper if and only if X is compact. 8.4.13 Exercise. Let X and Y be topological spaces. Show that projY : X × Y −→ Y is proper if and only if X is compact.
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8 Compactness
8.4.14 Exercise. Let f : X −→ Y and g : Y −→ Z be continuous maps. Prove (a) idX : X −→ X is proper and, if f and g are proper, then also g ◦ f is proper. (b) If g ◦ f is proper and f is surjective, then g is proper. (c) If g ◦ f is proper and g is injective, then f is proper. (d) If g ◦ f is proper and Y is Hausdorff, then f is proper. 8.4.15 Exercise. Assume that f : X −→ Y is a continuous map, X is a compact space, and Y is a Hausdorff space. Show that f is proper. 8.4.16 Exercise. Show that the inclusion (0, 1) ,→ R is not proper. (Cf. 8.4.1(b).) 8.4.17 Exercise. Let pk : C −→ C be given by pk (z) = z k , k ∈ Z. Is this map proper? 8.4.18 Exercise. Let X and Y be ininite discrete spaces. Under what conditions is a map f : X −→ Y proper? 8.4.19 Exercise. A diagram of topological spaces and continuous maps G
F
/E p
q
²
²
Y
f
/X
is called square!cartesian or pullback diagram if G = {(y, e) ∈ Y × E  f (y) = p(e)} (with the relative topology induced by the product topology) and q(y, e) = y, F (y, e) = F . Frequently G is denoted by Y ×X E and is called fibered product of Y and E over X. (See 5.3.6.) If X is a Hausdorff space, prove the following statements: (a) If p is proper, then q is proper. (b) If E is compact and f is proper, then G is compact. (Hint: Notice that since X is Hausdorff, the diagonal ∆X = {(x, x)  x ∈ X} ⊂ X × X is closed.)
195
8.5 Compactopen topology
8.5
Compactopen topology
Let X and Y be topological spaces and take Top(X, Y ) = {f : X −→ Y  f is continuous}. In this section we shall see how to furnish Top(X, Y ) with a convenient topology. A natural function is e : Top(X, Y ) × X −→ Y , given by e(f, x) = f (x), which is called evaluation. If we provide Top(X, Y ) with the discrete topology, then e is continuous. Which is the coarsest topology on Top(X, Y ) such that the evaluation e is continuous? To face this question we formulate another one, which is more general. Let X and Y be topological spaces and let Z be a set. Given a function of sets α : Z × X −→ Y , which topologies on Z make α continuous? 8.5.1 Lemma. Assume that α : Z × X −→ Y is continuous and let Nx be the neighborhood filter of x in X. Then (a) The map αz : X −→ Y given by αz (x) = α(z, x), is continuous for all z. (b) If F is a filter in Z such that F → z0 , then α(F × Nx ) → α(z0 , x) for all x ∈ X, where F × Nx is the filter with the set of products of an element of F and an element of Nx as filter basis. Proof: (a) This was already proved in Section 4.3. (b) It is clear that F × Nx → (z0 , x) (see 7.5.6). Since α is continuous we get the assertion. t u 8.5.2 Lemma. Take H ⊂ Top(X, Y ) and f0 ∈ H, and let F be a filter in H such that e(F × Nx ) → f0 (x) for all x ∈ X. Then there exists a topology on H for which F is the neighborhood filter of f0 and therefore F → f0 , and for which e is continuous. Proof: Define a topology A on H by A = {A ⊂ H  f0 6∈ A} ∪ {A ∈ F  f0 ∈ A} . Clearly this topology is such that its neighborhood filter at f0 in H, NfH0 , is contained in F, namely F → f0 .
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8 Compactness
We now see that e is continuous in topology. Take f ∈ H, f 6= f0 . For x ∈ X take a neighborhood Q of f (x) in Y . Since f : X −→ Y is continuous, f −1 (Q) is a neighborhood of x in X and {f } × f −1 (Q) is a neighborhood of (f, x) in H × X. Moreover e({f } × f −1 (Q)) = f f −1 (Q) ⊂ Q. Therefore e is continuous at (f, x) if f 6= f0 . The evaluation map e is also continuous at (f0 , x). Namely, by assumption e(F × Nx ) → f0 (x), we have that e maps the neighborhood filter of (f0 , x) to a convergent filter e(f0 , x) = f0 (x). t u Take H ⊂ Top(X, Y ), f0 ∈ H, and a filter F in H. Assume that H has a topology for which F → f0 . Then F → f0 for any coarser topology than the given one. Thus we have the following. 8.5.3 Proposition. Assume there is a coarsest topology on H for which e is continuous. In this topology, F → f0 if and only if e(F × Nx ) → f0 (x) for all x ∈ X. t u 8.5.4 Theorem. If Top(X, Y ) has a coarsest topology for which the evaluation is continuous, then a function f : Z × X −→ Y is continuous if and only if the following conditions hold: (a) For each point z ∈ Z, the function fz : X −→ Y , given by fz (x) = f (z, x), is continuous. (b) The function fe : Z −→ Top(X, Y ), given by fe(z) = fz , is continuous. Proof: We start showing that the conditions are sufficient. By (a) we can decompose f as the following diagram shows. Z × XP
f
PPP PPP PPP PP( e f ×id
/ pp8 Y p p pp pppe p p p
Top(X, Y ) × X .
By (b), fe × id is continuous, and thus f is continuous too. The two conditions are necessary. By Lemma 8.5.1(a), if f is continuous, so is fz too. Therefore (a) holds. To check (b), let G be a filter converging to z. Once more, Lemma 8.5.1(b) implies that for each point x ∈ X, f (G×Nz ) → f (z, x), namely e(fe×id)(G×Nz ) → f (z, x) = e(fe(z), x). But (fe × id)(G × Nz ) = fe(G) × Nx , so that, by Proposition 8.5.3, fe(G) → fe(z). Hence fe is continuous. t u
8.5 Compactopen topology
197
Take H ⊂ Top(X, Y ) and assume that H has a topology such that the evaluation e : H × X −→ Y is continuous. Let K ⊂ X be a compact set, let Q ⊂ Y be an open set, and define QK = {f ∈ H  f (K) ⊂ Q}, namely f ∈ QK if and only if e({f } × K) ⊂ Q. 8.5.5 Proposition. QK ⊂ H is an open set. Proof: Take f ∈ QK . For each point x ∈ K, Q is neighborhood de f (x) = e(f, x). Since e is continuous, there exist neighborhoods W (f, x) of f in H and V (f, x) of x in X such that e(W (f, x) × V (f, x)) ⊂ Q. Since K is compact, one can cover it with finitely many of the neighborhoods V (f, x). Say V (f, x1 ), . . . V (f, xn ) cover K. Thus W = W (f, x1 ) ∩ · · · ∩ W (f, xn ) is a neighborhood of f in H and e(W × V (f, xi )) ⊂ Q, for i = 1, . . . , n. Since K ⊂ V (f, x1 ) ∪ · · · ∪ V (f, xn ), one has e(W × K) ⊂ Q, i.e. W ⊂ QK . This shows that QK is open. t u We have shown that if a topology on H is such that the evaluation map e is continuous, then the sets QK must be open in this topology. Thus even the coarsest topology on H such that e is continuous must contain these sets. This suggests the following. 8.5.6 Definition. Take H ⊂ Top(X, Y ). The topology in H that has the family {QK  K ⊂ X is compact and Q ⊂ Y is open} as a subbasis, is called the compactopen topology in H. We shall use the notation M(X, Y ) for H = Top(X, Y ) with this topology. 8.5.7 Note. The sets QK do not form in general a basis for the topology of M(X, Y ). By construction of the compactopen topology we have the following. 8.5.8 Lemma. The compactopen topology in H ⊂ M(X, Y ) is coarser than any topology for which the evaluation e : H × X −→ Y is continuous. t u Let X be locally compact. Take (f0 , x0 ) ∈ H ×X, and an open neighborhood Q of f0 (x0 ) in Y . Then by the continuity of f0 , there exists a compact neighborhood K of x in X such that f0 (K) ⊂ Q. Hence f0 ∈ QK , and thus QK is a neighborhood of f0 in the compactopen topology. Hence QK × K is a neighborhood of (f0 , x0 ) and by definition, e(QK × K) ⊂ Q. Therefore e : H × X −→ Y is continuous with respect to the compactopen topology en H. By the previous lemma, we have shown the next.
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8.5.9 Theorem. Let X be a locally compact space and let Y be any space. Then for any H ⊂ M(X, Y ), the compactopen topology in H is the coarsest for which the evaluation e : H × X −→ Y is continuous. t u 8.5.10 Theorem. Let X be a locally compact space and let Y be any space. Take Y 0 = {f ∈ M(X, Y )  f is constant}, then there is a homeomorphism Y ≈ Y 0 . Moreover, M(X, Y ) is a Hausdorff space if and only if Y is Hausdorff too. Proof: The map i : Y −→ M(X, Y ) given by i(y) : x 7→ y for all x ∈ X, is an embedding whose image is Y 0 . Its inverse is given by f 7→ f (x) = e(f, x) for some x ∈ X. Therefore, if M(X, Y ) is a Hausdorff space, then Y 0 is a Hausdorff space and thereby Y is also a Hausdorff space. Conversely, if Y is a Hausdorff space, take f, g ∈ M(X, Y ), f 6= g. Therefore, for some x ∈ X, one has f (x) 6= g(x). Thus there are open neighborhoods Q1 and {x} Q2 of f (x) and g(x) in Y , respectively such that Q1 ∩ Q2 = ∅. Hence the sets Q1 {x} {x} {x} and Q2 are open and disjoint, since if h ∈ Q1 ∩ Q2 , h(x) ∈ Q1 ∩ Q2 , which is {x} {x} impossible. Clearly f ∈ Q1 and g ∈ Q2 . Hence M(X, Y ) is a Hausdorff space. t u To finish this section, we shall give a very interesting application of the compactopen topology, that occurs when the domain space is locally compact. First we notice in the next example, due to Dieudonn´e, that in general the product of identifications is not necessarily an identification. 8.5.11 Example. Let Q denote the set of rational numbers with the relative topology and take relation ∼ on Q that identifies all integers in one point. Let p : Q −→ Q/ ∼ be the quotient map, which is an identification. However the product map p × id : Q × Q −→ Q/∼ ×Q is not an identification. 8.5.12 Exercise. Prove that in the previous example, the identification p is not an open map, but it is indeed a closed map. Moreover prove that in fact the product map p × idQ is not an identification. The next result, due to J. H. C. Whitehead, is a strong application of compactness. It solves the problem of when the product of identifications is again an identification. As we saw in Example 8.5.11, this in general not true.
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8.6 The exponential law
8.5.13 Theorem. Let p : X −→ X 0 be an identification and let Y be a locally compact space. Then the product map p × id : X × Y −→ X 0 × Y is an identification. Proof: Take an equivalence relation in X × Y given by (x, y) ∼ (x0 , y) if and only if p(x) = p(x0 ) and let q : X × Y −→ Z = X × Y /∼ be the quotient map. By the universal property of the quotient maps and since p × id and q identify the same points, there exists a unique continuous bijective map h : Z = X × Y /∼−→ X 0 × Y . By 8.5.4(b) and 8.5.9, q determines a map qe : X −→ M(Y, Z), continuous with respect to the compactopen topology on M(Y, Z). Moreover, if p(x) = p(x0 ), then clearly qe(x) = qe(x0 ). Thus, since p is an identification, one has a continuous map q 0 : X 0 −→ M(Y, Z) such that q 0 ◦ p = qe . Since the evaluation e : M(Y, Z)×Y −→ Z is continuous and Y is locally compact, the function e ◦ q 0 : X 0 × Y −→ Z, which coincides with h−1 , is continuous. Hence h is a homeomorphism and thus p × id is also an identification. t u In Section 8.7 (8.7.21) below we shall state a generalization of this result.
8.6
The exponential law
Given arbitrary sets X and Y , we denote (provisionally) by X Y the set of functions f : X −→ Y . If X, Y, Z are sets, then the exponential law establishes an equivalence of sets Z X×Y ∼ = (Z Y )X . For these, we only have to define ϕ : Z X×Y → (Z Y )X by ϕ(f )(x)(y) = f (x, y) and, as an inverse, ψ : (Z Y )X → Z X×Y by ψ(g)(x, y) = g(x)(y). Now we try to establish an analogous result for the space M(X, Y ), if X and Y are topological spaces. 8.6.1 Proposition. Let X, Y, Z topological spaces such that Y is Hausdorff and locally compacto. Then there is a setequivalence ϕ : M(X × Y, Z) → M(X, M(Y, Z)) .
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8 Compactness
Proof: To define ϕ, as we did at the beginning of the section, we have to prove that if f : X × Y → Z is continuous, then ϕ(f )(x) : Y → Z is continuous and that ϕ(f ) : X → M(Y, Z) is also continuous. For the first statement, observe that ϕ(f )(x) is the composite i
f
x Y −→X × Y −→Z ,
where ix (y) = (x, y), which is clearly continuous. (Notice that if X = ∅, then the result is trivial.) For the second statement, let U K be a subbasic open set of M(Y, Z). It is enough to see that ϕ(f )−1 (U K ) is open in X. Thus take x ∈ ϕ(f )−1 (U K ). Then ϕ(f )(x)(y) = f (x, y) ∈ U for all y ∈ K and there are neighborhoods Wy of x and Vy of y such that f (Wy × Vy ) ⊂ U . Since K is compact, the family {Vy } contains a finite subfamily V1 , . . . , Vm which covers K. Take W = W1 ∪ · · · ∪ Wm , where Wi is such that f (Wi × Vi ) ⊂ U . W is a neighborhood of x in X. Let us see now that W ⊂ ϕ(f )−1 (U K ). Indeed, if we take x0 ∈ W and y ∈ K, then ϕ(f )(x0 )(y) = f (x0 , y), but one has y ∈ Vi for some i, and x0 ∈ Wi , hence f (x0 , y) ∈ U . We have thus shown that ϕ is well defined. Let us see now that with the definition given at the beginning of the section, the map ψ : M(X, M(Y, Z)) → M(X × Y, Z) is well defined. Take g : X → M(Y, Z) continuous. It is enough to show that ψ(g) is continuous. To see this, take an open set U ⊂ Z and we show that ψ(g)−1 (U ) is open. Take (x, y) ∈ ψ(g)−1 (U ), i.e. g(x)(y) ∈ U . Since g(x) is continuous, there is a neighborhood W of y such that g(x)(W ) ⊂ U . Since Y locally compact and Hausdorff, there is a compact neighborhood K such that y ∈ K ⊂ W Hence g(x)(K) ⊂ U and therefore g(x) ∈ U K ), which is open in M(Y, Z). Since g is continuous, there is a neighborhood V of x in X such that g(V ) ⊂ U K . V × K is a neighborhood of (x, y) en X × Y . Take (x0 , y 0 ) ∈ V × K. Then ψ(g)(x0 , y 0 ) = g(x0 )(y 0 ) ∈ U and thus V × K ⊂ ψ(g)−1 (U ). t u With an additional condition the setequivalence in the previous result is a homeomorphism. I.e. we obtain the exponential law. 8.6.2 Theorem. If X, Y , Z ar topological spaces such that X and Y are Hausdorff spaces and Y is locally compact, then ϕ : M(X × Y, Z) → M(X, M (Y, Z))
201
8.6 The exponential law
is a homeomorphism. Proof: We see that ϕ and ψ are continuous. First take a subbasic set (U L )K in M(X, M(Y, Z)) with U open in Z and K and L compact in X and Y , respectively. Then K ×L is compact, and if f ∈ (U K×L ) ⊂ M(X × Y, Z), then ϕ(f )(K)(L) = f (K × L) ∈ U  I.e. ϕ(U K×L ) ⊂ (U L )K . Let now U J be a subbasic set in M(X × Y, Z), with J compact in X × Y . Take K = projX (J) and L = projY (J). K and L are compact and J ⊂ K × L. Let us see ψ((U L )K ) ⊂ U J . Indeed, take g ∈ (U L )K and (x, y) ∈ J, then ψ(g)(x, y) = g(x)(y) ∈ U , since x ∈ K and y ∈ L. t u We have the map (8.6.3)
M(X, Y ) × M(Y, Z) → M(X, Z)
given by composition. 8.6.4 Exercise. Prove that if X and Y are Hausdorff locally compact spaces, then the map (8.6.3) is continuous. In particular, if f : X → Y is continuous, then (by restriction of (8.6.3)) it induces a continuous map f # : M(Y, Z) → M(X, Z) , given by f # (g) = g ◦ f . Similarly, if g : Y → Z is continuous, then it induces (again by restriction of (8.6.3)) a continuous map g# : M(X, Y ) → M(X, Z) such that g# (f ) = g ◦ f . 8.6.5 Definition. Let A be a subspace of X and B a subspace of Y . Denote by M(X, A; Y, B) the subspace of M(X, Y ) consisting of the maps f : X → Y such that f (A) ⊂ B. An important instance of these subspaces is the space M(X, x0 ; Y, y0 ) of maps f : X → Y such that f (x0 ) = y0 , if x0 ∈ X and y0 ∈ Y are specified points. These maps are called pointed maps (or (o based maps), since they map the base point x0 of X to the base point y0 of Y . The pairs (X, x0 ) or (Y, y0 ) are called pointed spaces. 8.6.6 Example. Let I = [0, 1] be the interval and let ∂I = {0, 1} be its boundary. Thus we can consider the spaces M(I, X) ⊃ M(I, 0; X, x0 ) ⊃ M(I, ∂I; X, x0 ) ,
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8 Compactness
for a pointed space (X, x0 ). These spaces are known as free pathspace in X, pathspace in X based in x0 , and loopspace of X based at x0 , respectively. The set M(I, ∂I; X, x0 ) is usually denoted by Ω(X, x0 ) or if the base point is clear, then by ΩX (compare with 8.6.9). 8.6.7 Definition. Let us consider two pairs of spaces (X, A) and (Y, B). We define its product as the pair of spaces (X, A) × (Y, B) = (X × Y, X × B ∪ A × Y ) . Hence (I, ∂I) × (I, ∂I) = (I 2 , ∂I 2 ), where I 2 is the unit space in the plane and ∂I 2 its boundary, which is homeomorphic to the circle S1 (see Figure 8.2).
≈
I ×I
I
B2
I
I × ∂I
∂I
I
∪
I
∂I × I
=
∂(I × I)
≈
S1
∂I
Figure 8.2 (I, ∂I) × (I, ∂I) = (I 2 , ∂I 2 ) ≈ (B2 , S1 )
Inductively, (I n , ∂I n ) × (I, ∂I) = (I n+1 , ∂I n+1 ), where I n+1 is the unit cube in and ∂I n+1 is its boundary, que is homeomorphic to the sphere
Rn+1
Sn = {(x1 , . . . , xn+1 ) ∈ Rn+1  x21 + · · · + x2n+1 = 1}. By the exponential law (which is also valid for pairs –exercise), we have (8.6.8)
M(I n+1, ∂I n+1 ; X, x0 )≈M(I, ∂I;M(I n, ∂I n ; X, x0 ), x f0 ) ,
where x f0 ∈ M(I n , ∂I n ; X, x0 ) is such that x f0 (I n ) = x0 . 8.6.9 Definition. The space M(I n , ∂I n ; X, x0 ) is called the nloop space in X based in x0 and and is denoted by Ωn (X, x0 ) .
8.7 Compactly generated spaces
203
If the base point is clear, then we abuse the notation and write Ωn X. By (8.6.8) we have Ω(Ωn (X, x0 ), x f0 ) ≈ Ωn+1 (X, x0 ). 8.6.10 Exercise. Let X be a pointed space. Show that there is a homeomorphism Ωn (X, x0 ) ≈ M(Sn , ∗; Z, x0 ).
8.7
Compactly generated spaces
For several constructions of new topological spaces out of old ones, it it is convenient to have an adequate topology on the spaces, in order to have better properties. In this section we shall study a certain class of spaces, whose topology is determined by their compact subsets. We shall also how to modify a given topology on any Hausdorff space so that it becomes a space with this new topology. This topology does not alter very much the given topology, so that using it we may deduce many properties of the original one, particularly homotopy theoretical properties of the space. This topology was introduced by Kelley [12] and was studied in detail by Steenrod [18], whose ideas we partly follow. Other references for properties of this class of spaces are [4] and [5]. There are other convenient classes of topological spaces which are similar to but more general than this one. At the end of the section we shall say some words about them. 8.7.1 Definition. A Hausdorff space X is said to be compactly generated if the following axiom holds: (CG) A subset A ⊂ X is closed if and only if A ∩ K is closed for any compact subset K ⊂ X. In other words, a space is compactly generated if and only if it has the weak topology generated by its compact subspaces, i.e. the finest topology for which the inclusions K ,→ X are continuous for all compact subsets K ⊂ X. I.e. we have the following universal property. 8.7.2 Theorem. Let X be a Hausdorff space. Then X is compactly generated if and only if it satisfies the following condition: For every space Y and any map f : X −→ Y , f is continuous if and only if f K is continuous for each compact subset K ⊂ X.
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8 Compactness
Proof: Assume X is compactly generated and take a map f : X −→ Y . If f is continuous, then f K Is continuous too for every compact set K. Conversely, if f K is continuous para every compact subset K ⊂ X and D ⊂ Y is a closed set, then (f −1 D) ∩ K = (f K )−1 (D) is closed in K. Hence, since X is compactly generated, f −1 D is closed and thus f is continuous. Assume now that X is a Hausdorff space and satisfies the condition, and take e be the A ⊂ X such that for each compact set K ⊂ X, A ∩ K is closed. Let X space with the same underlying set as X, but its closed sets are precisely those sets B such that B ∩ K is closed in X for each compact subset K ⊂ X. Hence e On the other hand, by the very definition of X, e the identity A is closed in X. e is such that idK is continuous. Hence by the condition id map id : X −→ X is continuous and therefore A = id−1 A is closed in X. Thus X is compactly generated. t u e as defined in the previous proof, is indeed a topo8.7.3 Exercise. Prove that X, logical space, namely, that the set of its closed sets satisfies the axioms for closed e is a compactly generated space. (Hint: Check sets 2.2.5. Moreover prove that X e and X have the same compact sets.) that X The next result gives a criterion, i.e. a sufficient condition for a space to be compactly generated. 8.7.4 Theorem. Let X be a Hausdorff space such that for every subset A ⊂ X and for each point x ∈ A−A, there is a compact set K ⊂ X such that x ∈ A ∩ K−A∩K. Then X is compactly generated. Proof: Let A ⊂ X be such that for every compact set K ⊂ X, the intersection A ∩ K is closed, and let x be a cluster point of A. By assumption there is a compact set K0 ⊂ X such that x is a cluster point of A ∩ K0 . Since this set is closed, x ∈ A ∩ K0 ⊂ A. Thus A is closed. Hence X is compactly generated. t u The converse of the previous result is false, as the following example shows. 8.7.5 Example. Let Y be the space whose points are all ordinals less than or equal to Ω, the first uncountable ordinal. Let Y have the order topology, i.e. the topology which has the half lines Ia = {y ∈ Y  y < a} and I b = {y ∈ Y  b < y}, a, b ∈ Y , as subbasis. Now take as X the subspace of Y obtained by deleting all limit ordinals (i.e. those ordinals without an immediate predecessor), except Ω. Thus K ⊂ X is compact if and only if K is finite, since every infinite
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8.7 Compactly generated spaces
subset of X contains a sequence that converges to a limit ordinal different from Ω. Consequently A = X − {Ω} meets each compact set in a closed set, However A is not closed, since Ω ∈ A. Therefore X is not compactly generated. On the other hand Y is compactly generated. It is indeed compact. This shows that not every (open) subspace of a compactly generated space is compactly generated. Before we analyze when the property of a space being compactly generated is inherited by subspaces, we shall see that the class of compactly generated spaces is ample, so that many of the spaces that come from applications of topology are of this sort. We shall do this using criterion 8.7.4. 8.7.6 Proposition. Every locally compact Hausdorff space X is compactly generated. Proof: Let x ∈ X be a cluster point of a set A ⊂ X and teke a compact neighborhood K of x. Then the intersection A ∩ K 6= ∅ and x is a cluster point of it. By 8.7.4, X is compactly generated. t u 8.7.7 Theorem. Let X be a Hausdorff space. Then X is compactly generated if and only if X is a quotient of a locally compact space. Proof: Assume that X is compactly generated. Since it has the weak topology induced by its compact subspaces, it is a quotient of the topological sum of all of them. Namely, a q:Y = K −→ X , K⊂X , K compact
given by the inclusions qK = iK : K ,→ X, is an identification. Obviously Y is locally compact. Conversely, if there is an identification q : Y −→ X with Y locally compact, take A ⊂ X such that A ∩ C is open in K for each compact set K ⊂ X. We shall see that A is open en X. If V ⊂ Y is an open set such that V is compact, then A ∩ q(V ) = q(V ) ∩ G for some open set G in X. Since q −1 (A) ∩ q −1 q(V ) = q −1 q(V ) ∩ q −1 (G), if we intersect it with V , then we have that q −1 (A) ∩ V = V ∩ q −1 (G). Hence q −1 (A) ∩ V is open in Y . Since Y is locally compact, we can cover it with a family {Vα } of relatively compact open sets, in such a way that S q −1 (A) = α (q −1 (A) ∩ Vα ). This shows that q −1 (A) is open in Y and since q is an identification, A is open en X. t u Similarly to 8.7.6 we have the following.
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8 Compactness
8.7.8 Proposition. Every firstcountable Hausdorff space is compactly generated. Proof: Let X be firstcountable and let x be a cluster point of a set A ⊂ X. Hence there is sequence xn ∈ A such that xn → x. Put K = {xn } ∪ {x}. Hence the intersection A ∩ K 6= ∅ and x is a cluster point of it. By 8.7.4, X is compactly generated. t u We shall see now that the property of being compactly generated is inherited by the closed subspaces and by some open subspaces. 8.7.9 Definition. Let X be a topological space and take Y ⊂ X. We say that Y is a regular open set if every point y ∈ Y has a neighborhood V in X such that V ⊂Y. 8.7.10 Proposition. Let X be a compactly generated space and take Y ⊂ X. If Y is closed or regular open, then Y , furnished with the relative topology, is compactly generated. Proof: Assume that Y is closed and take B ⊂ Y such that B ∩ K is closed for every compact set K ⊂ Y . If K 0 ⊂ X is compact, then B ∩ K 0 = B ∩ (K 0 ∩ Y ) is closed since (K 0 ∩ Y ) ⊂ Y is compact. Therefore B is closed in X and thus it is closed in Y too. Assume now that Y is regular open and take B ⊂ Y such that B ∩ K is closed in Y for every compact set K ⊂ Y . Let y be a cluster point of B in Y . Hence there is a neighborhood V of y in X such that V ⊂ Y . If K 0 ⊂ X is compact, then V ∩ K 0 ⊂ Y is compact. Consequently B ∩ V ∩ K 0 is closed in Y , is closed in V ∩ K 0 and hence also in X. Therefore, since K 0 is arbitrary and X is compactly generated, we have that B ∩ V is closed in X. Finally, since y is a cluster point of B, it is also a cluster point of B ∩ V . Hence, since B ∩ V is closed, y ∈ B ∩ V ⊂ B, i.e. y ∈ B. Hence B is closed. t u 8.7.11 Exercise. Show that the definition of regular open set given above coincides with the one given in Exercise 2.2.22. The assumption that Y is regular open in X in the previous proposition cannot be replaced by the assumption that Y is open, as we showed in Example 8.7.5. Dually we can ask if an identification space of a compactly generated space is a compactly generated space. First we notice that there is a necessary condition for it is the fact that the resulting space has to be Hausdorff, which is not always the case. Indeed this condition is also sufficient.
8.7 Compactly generated spaces
207
8.7.12 Proposition. Let X be a compactly generated space and let q : X −→ Z be an identification such that Z is Hausdorff. Then Z is compactly generated. Proof: Since X compactly generated, by 8.7.7 there is a locally compact space Y and an identification p : Y −→ X. Hence the composite q ◦ p : Y −→ Z is identification too, and again by 8.7.7, Z is compactly generated. t u Given any Hausdorff space X, as we saw in the proof of 8.7.2, there is a space e which is compactly generated. associated to X, which we called X, 8.7.13 Definition. Let X be a Hausdorff space. Define c(X) as the space with the same underlying space as X but furnished with the topology whose family of closed sets is C = {A ⊂ X  A ∩ K is closed in X for all compact subsets K ⊂ X} . As it was asked in Exercise 8.7.3, this family C satisfies the axioms for the closed sets of topological space, and thus it makes c(X) a compactly generated space which has exactly the same compact sets of X. We call c(X) the compactly generated space associated to X . 8.7.14 Remark. Usually what we denote by c(X) is denoted by k(X). However we avoid this notation, since k(X) will denote a similar but more general construction, that we present below in 8.8.1. This construction is functorial, namely, given any map f : X −→ Y , the same function between sets determines a continuous map c(f ) : c(X) −→ c(Y ) with the following properties: (a) (b)
c(idX ) = idc(X) : c(X) −→ c(X), if X is a Hausdorff space, and c(g ◦ f ) = c(g) ◦ c(f ) : c(X) −→ c(Z), if f : X −→ Y and g : Y −→ Z are continuous maps between Hausdorff spaces.
8.7.15 Exercise. Show that indeed c(f ) : c(X) −→ c(Y ) is continuous. The following result summarizes the fundamental properties of c(X).
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8 Compactness
8.7.16 Theorem. Let X and Y be Hausdorff spaces. Then c(X) has the following properties. (a) The identity map j : c(X) −→ X is continuous. (b) c(X) is a Hausdorff space. (c) X and c(X) have the same compact sets. (d) c(X) is compactly generated. (e) If X is compactly generated, then c(X) = X. (f) If f : X −→ Y is continuous on each compact subspace of X, then c(f ) : c(X) −→ c(Y ) is continuous. Proof: (a) It is clear since any closed set in X is also closed in c(X). (b) It is clear too since the neighborhoods in X are neighborhoods also in c(X) (the topology of c(X) is finer than that of X). (c) If K ⊂ c(X) is compact, then by (a) K ⊂ X is compact too (it is the image e ⊂ c(X) under j of a compact set). Conversely, let K ⊂ X be compact and let K e −→ K is be the same set K with the relative topology. By (a), the identity K continuous. To see that its inverse map is also continuous, let B ⊂ K be closed. Obviously, B meets each compact subset of X in a closed set. Therefore B is e This proves that the identity X −→ X e is closed in c(X) and hence also in K. e is compact. continuous. Hence X (d) If A meets each compact subset of c(X) in a closed set, then by (c) A meets any compact subset of X in a compact set, thus also in a closed set. Thus A is closed in c(X). (e) It follows directly from (d). (f) Let B ⊂ c(Y ) be a closed set. Then B ∩ L is closed in Y for each compact set L. Let K ⊂ X be compact. Then f −1 (B) ∩ K = (f K )−1 (B ∩ f (K)) is closed in K and hence also in X, since f (K) ⊂ Y is compact. This proves that f −1 (B) is closed in c(X), and thus c(f ) : c(X) −→ c(Y ) is continuous. t u There is a universal property that que characterizes the construction c(X). 8.7.17 Proposition. Let X be a Hausdorff space. The identity map c(X) −→ X has the following universal property that characterizes it.
209
8.7 Compactly generated spaces
(CG) Let Y be a compactly generated space and f : Y −→ X a continuous map, then there is a unique continuous map fb : Y −→ c(X) such that the following diagram commutes: / c(X) 0}. The sets A and B are closed and disjoint, however their distance is µ(A, B) = 0, as one easily checks. The next result is clear. 9.1.15 Proposition. If A is nonempty, then µ(x, A) = 0 if and only if x ∈ A. t u 9.1.16 Theorem. Sea X un metric space with metric d and take a nonempty subset A ⊂ X. Then the function δA : X −→ R given by δA (x) = µ(x, A) is continuous. Proof: For x, y ∈ X, z ∈ A, the triangle inequality tells us d(x, z) ≤ d(x, y) + d(y, z) . Therefore, for each z ∈ A, one has µ(x, A) ≤ d(x, y) + d(y, z) , so that µ(x, A) ≤ d(x, y) + µ(y, A) . This shows the inequality δA (x) − δA (y) = µ(x, A) − µ(y, A) ≤ d(x, y) , which clearly implies the continuity of δA .
t u
From 9.1.15 and 9.1.16, we obtain the next statement. 9.1.17 Proposition. Take a nonempty set A ⊂ X, then A = ∩∞ n=1 Qn , where n Qn = {x ∈ X  µ(x, A) < 1/2 }, which is an open set. In particular, every closed set is a countable intersection of open sets and, consequently, every open set is a countable union of closed sets. t u
221
9.1 Normal spaces
Since δA is a continuous realvalued function, we obtain the next result from Corollary 8.1.32. 9.1.18 Corollary. Take a nonempty closed set A ⊂ X and a nonempty compact set B ⊂ X. If A ∩ B = ∅, then µ(A, B) > 0. t u The main theorem of this section is the following. 9.1.19 Theorem. Every metric space is a T4 space. Proof: Let A, B ⊂ X be two nonemepty disjoint closed sets and let h : X −→ R be given by h(x) = µ(x, A) − µ(x, B) = δA (x) − δB (x). Clearly, h is a continuous function so that we have open sets U = {x  h(x) < 0}
and
V = {x  h(x) > 0} ,
which are obviously disjoint. Obviously, if x ∈ A, then µ(x, B) > 0 (otherwise x ∈ B = B), so that h(x) < 0, namely A ⊂ U . Analogously B ⊂ V . This shows that it is possible to separate A and B by neighborhoods. t u 9.1.20 Note. Every subspace of a metric space is metric and therefore a T4 space. 9.1.21 Definition. A normal space X such that every subspace of X is normal, is called completely normal. Therefore, every metric space is completely normal. 9.1.22 Lemma. Axiom (N) is equivalent to (N’) For every closed subset B ⊂ X and every open subset Q ⊂ X such that B ⊂ Q, there is an open set U ⊂ X such that B ⊂ U ⊂ U ⊂ Q. One such open set U is called a shrinking of the neighborhood Q de B. Proof: (N) =⇒ (N’) Let B be closed in X and let Q be an open neighborhood of B. Then B and X − Q are disjoint closed sets and by (N) there are disjoint open neighborhoods U of B and V of X − Q, namely U ⊂ X − V . The set X − V is closed. Therefore, B ⊂ U ⊂ U ⊂ X − V ⊂ X − (X − Q) = Q. (N’) =⇒ (N) Take disjoint open sets A, B ⊂ X. Therefore, Q = X − A is an open neighborhood of B. By (N’) there is an open neighborhood U of B such that U ⊂ Q. Consequently, X − U ⊃ X − Q = A, and U and X − U are disjoint neighborhoods of B and A, respectively. t u
222
9 Other separability axioms
The repeated application of the previous lemma produces a very interesting construction. Namely, if we take disjoint closed sets A and B in a space X that satisfies (N), then take U (1) = X − A , which is an open neighborhood of B. Applying the lemma, there is an open neighborhood U (0) of B such that U (0) ⊂ U (1) . If we continue this process, we obtain open sets U ( 12 ), U ( 41 ), U ( 34 ) such that U (0) ⊂ U ( 14 ) ⊂ U ( 14 ) ⊂ U ( 12 ) ⊂ U ( 21 ) ⊂ U ( 34 ) ⊂ U ( 34 ) ⊂ U (1) . Continuing this process of putting a new neighborhood between every two already obtained, we get inductively, for each k/2n , k = 0, . . . , 2n , open sets U ( 2kn ) such that (7.1.23)
U ( 2kn ) ⊂ U ( k+1 2n )
k = 0, . . . , 2n − 1 ,
since given U ( 2kn ) and applying the lemma to (7.1.23), we obtain U ( 2k+1 ) such 2n+1 that U ( 2kn ) ⊂ U ( 2k+1 ) ⊂ U ( 2k+1 ) ⊂ U ( k+1 2n ) . 2n+1 2n+1 This is the family of neighborhoods that corresponds to the stage n + 1. Thus we have obtained for all r ∈ [0, 1] such that r = that r < r0 ⇒ U (r) ⊂ U (r0 ) .
k 2n ,
open sets U (r) such
Now, for any t ∈ [0, 1] we define the open set [ U (t) = U (r) r≤t
and one has t < t0 ⇒ U (t) ⊂ U (t0 ) ,
(9.1.24)
since if t < t0 , then there are k and n such that t < U (t) ⊂
U ( 2kn )
⊂
U ( 2kn )
⊂
k 2n
U ( k+1 2n )
k+1 0 2n < t . U (t0 ) .
1. Then (9.1.24) is still valid. Define f (x) = inf{t ∈ R  x ∈ U (t)}. Since U (t) = X for t > 1, then f (x) ≤ 1 if x ∈ X, and since U (t) = ∅ for t < 0, then f (x) ≥ 0 for all x ∈ X. Moreover, if B 6= ∅, then f (x) = 0 for all x ∈ B and if A 6= ∅, then f (x) = 1 for all x ∈ A.
9.1 Normal spaces
223
We shall now show that the function f is continuous. Take x0 ∈ X and ε > 0. Put t0 = f (x0 ). We shall prove that there is a neighborhood of x0 such that f (x) − t0  ≤ ε for all x in this neighborhood. If x ∈ U (t0 + ε), then f (x) ≤ t0 + ε. Moreover, if x ∈ X − U (t0 − ε), then f (x) ≥ t0 − ε (since if f (x) < t0 − ε, then x ∈ U (t0 − ε) ⊂ U (t0 − ε)). Thus if x ∈ V = U (t0 + ε) ∩ (X − U (t0 − ε)), then f (x) − t0  ≤ ε. Also V is open and x0 ∈ V since f (x0 ) = t0 . Therefore x0 ∈ U (t0 + ε) y x0 6∈ U (t0 − 2ε ) ⊃ U (t0 − ε). As a consequence of all the previous, we have that f is continuous. Thus, given disjoint closed sets A and B in X, we have constructed a continuous function f : X −→ [0, 1] such that f B = 0 y f A = 1. Conversely, given such a function f , we have that the sets f −1 [0, 1/2) and f −1 (1/2, 1] are open and disjoint and they contain B and A, respectively. Thus we have proved the following. 9.1.25 Theorem. (Urysohn’s lemma) In a topological space X axiom (N) (of Definition 9.1.4) and (N00 ) Given disjoint closed sets A, B ⊂ X, there is a continuous function f : X −→ [0, 1] such that f A ≡ 1 and f B ≡ 0. t u 9.1.26 Definition. A function f : X −→ [0, 1], as in (N00 ), is called Urysohn’s function and an open neighborhood of A ⊂ X (or an open set) of the form f −1 (0, 1] is called numerable neighborhood of A (numerable open set). 9.1.27 Note. If B = ∅, then we can take f ≡ 1. 9.1.28 Note. If A and B are disjoint closed sets in X and we consider Y = A∪B, then the function defined on Y ⊂ X with value 1 on A and 0 on B is continuous. The statement of (N00 ) means that this function can be extended to all of X. 9.1.29 Theorem. (Tietze extension theorem) In a topological space X axiom (N) is equivalent to (N000 ) If G is a closed set in X and g : G −→ [a, b] is a continuous function, then G can be continuously extended to a function f : X −→ [a, b]. 000
To prove (N ) one constructs a sequence of functions fn : X −→ [a, b] such that the restrictions fn G approximate more and more the function g and the sequence {fn (x)} converges uniformly. The induction step to construct fn+1 once we have fn is a consequence of the next lemma.
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9 Other separability axioms
9.1.30 Lemma. Let X be a normal space (that satisfies (N)) and take a closed set G in X and b > 0. If u : G −→ [−b, b] is continuous, then there is v : X −→ [− 3b , 3b ] such that for all x ∈ G, b u(x) − v(x) ≤ . 3 Proof: Take b b A = {x ∈ G  u(x) ≤ − } and B = {x ∈ G  u(x) ≥ }. 3 3 The sets A and B are closed and disjoint, so that there is a continuous function w : X −→ [0, 1] such that wA = 0 and wB = 1. Define v(x) = Then for all x, v(x) ≤
b 3
2b b w(x) − . 3 3
and if x ∈ G, then
u(x) − v(x) = u(x) +
b 2b − w(x). 3 3
Take x ∈ G. There are three possibilities, namely b (i) u(x) ≤ − . 3 b In this case x ∈ A, so that v(x) = − . Therefore, 3 −
2b b b b = −b + ≤ u(x) − v(x) ≤ − + = 0 . 3 3 3 3
(ii) u(x) ≥ 3b . In this case x ∈ B, so that v(x) = 3b . Therefore 2b b b b = b − ≥ u(x) − v(x) ≥ − = 0 . 3 3 3 3 b b (iii) − < u(x) < . 3 3 In this case − 3b ≤ v(x) ≤ 3b , so that −
b b b b 2b 2b = − − < u(x) − v(x) < + = . 3 3 3 3 3 3
In all three cases one has the desired inequality.
t u
225
9.1 Normal spaces
Proof de 9.1.29: 000 00 00 000 (N ) ⇒ (N ) is clear, and since (N ) ⇔ (N), we have that (N ) ⇒ (N). 000
The only assertion that we have to prove is (N) ⇒ (N ). Without loss of generality, we may assume that a = −1 and b = 1, namely that we have a function g : G −→ [−1, 1]. By the previous lemma there is a continuous function f1 : X −→ [− 13 , 13 ] such that 2 g(x) − f1 (x) ≤ . 3 2 2 Take g1 = g − f1 : G −→ [− 3 , 3 ]. Therefore, there is a continuous function v1 : X −→ [− 29 , 29 ] such that 2 g1 (x) − v1 (x) ≤ ( )2 . 3 Take f2 = f1 + v1 : X −→ [−1 + ( 23 )2 , 1 − ( 23 )2 ]. Assume that we have constructed a continuous function 2 2 fn : X −→ [−1 + ( )n , 1 − ( )n ] 3 3 such that g(x) − fn (x) ≤ ( 23 )n for x ∈ G. If we apply the previous lemma to gn = g − fn G , we have that there is another continuous function 1 2 1 2 vn : X −→ [− ( )n , ( )n ] 3 3 3 3 such that gn (x) − vn (x) < ( 32 )n+1 for x ∈ G. Now we define fn+1 (x) = fn (x) + vn (x). Clearly g(x) − fn+1 (x) < ( 23 )n+1 for x ∈ G and 2 2 fn+1 : X −→ [−1 + ( )n+1 , 1 − ( )n+1 ] . 3 3 2 n Since g(x) − fn (x) < ( 3 ) for all n, we get for x ∈ G that lim fn (x) = g(x). Take an arbitrary point x ∈ X and m ≥ n. Then fm (x) − fn (x) = 
m−1 X
vk (x)